I want to split a command line like string in single string parameters. How look the regular expression for it. The problem are that the parameters can be quoted. For example like:
"param 1" param2 "param 3"
should result in:
param 1, param2, param 3
You should not use regular expressions for this. Write a parser instead, or use one provided by your language.
I don't see why I get downvoted for this. This is how it could be done in Python:
>>> import shlex
>>> shlex.split('"param 1" param2 "param 3"')
['param 1', 'param2', 'param 3']
>>> shlex.split('"param 1" param2 "param 3')
Traceback (most recent call last):
[...]
ValueError: No closing quotation
>>> shlex.split('"param 1" param2 "param 3\\""')
['param 1', 'param2', 'param 3"']
Now tell me that wrecking your brain about how a regex will solve this problem is ever worth the hassle.
I tend to use regexlib for this kind of problem. If you go to: http://regexlib.com/ and search for "command line" you'll find three results which look like they are trying to solve this or similar problems - should be a good start.
This may work:
http://regexlib.com/Search.aspx?k=command+line&c=-1&m=-1&ps=20
("[^"]+"|[^\s"]+)
what i use
C++
#include <iostream>
#include <iterator>
#include <string>
#include <regex>
void foo()
{
std::string strArg = " \"par 1\" par2 par3 \"par 4\"";
std::regex word_regex( "(\"[^\"]+\"|[^\\s\"]+)" );
auto words_begin =
std::sregex_iterator(strArg.begin(), strArg.end(), word_regex);
auto words_end = std::sregex_iterator();
for (std::sregex_iterator i = words_begin; i != words_end; ++i)
{
std::smatch match = *i;
std::string match_str = match.str();
std::cout << match_str << '\n';
}
}
Output:
"par 1"
par2
par3
"par 4"
Without regard to implementation language, your regex might look something like this:
("[^"]*"|[^"]+)(\s+|$)
The first part "[^"]*" looks for a quoted string that doesn't contain embedded quotes, and the second part [^"]+ looks for a sequence of non-quote characters. The \s+ matches a separating sequence of spaces, and $ matches the end of the string.
Regex: /[\/-]?((\w+)(?:[=:]("[^"]+"|[^\s"]+))?)(?:\s+|$)/g
Sample: /P1="Long value" /P2=3 /P3=short PwithoutSwitch1=any PwithoutSwitch2
Such regex can parses the parameters list that built by rules:
Parameters are separates by spaces (one or more).
Parameter can contains switch symbol (/ or -).
Parameter consists from name and value that divided by symbol = or :.
Name can be set of alphanumerics and underscores.
Value can absent.
If value exists it can be the set of any symbols, but if it has the space then value should be quoted.
This regex has three groups:
the first group contains whole parameters without switch symbol,
the second group contains name only,
the third group contains value (if it exists) only.
For sample above:
Whole match: /P1="Long value"
Group#1: P1="Long value",
Group#2: P1,
Group#3: "Long value".
Whole match: /P2=3
Group#1: P2=3,
Group#2: P2,
Group#3: 3.
Whole match: /P3=short
Group#1: P3=short,
Group#2: P3,
Group#3: short.
Whole match: PwithoutSwitch1=any
Group#1: PwithoutSwitch1=any,
Group#2: PwithoutSwitch1,
Group#3: any.
Whole match: PwithoutSwitch2
Group#1: PwithoutSwitch2,
Group#2: PwithoutSwitch2,
Group#3: absent.
Most languages have other functions (either built-in or provided by a standard library) which will parse command lines far more easily than building your own regex, plus you know they'll do it accurately out of the box. If you edit your post to identify the language that you're using, I'm sure someone here will be able to point you at the one used in that language.
Regexes are very powerful tools and useful for a wide range of things, but there are also many problems for which they are not the best solution. This is one of them.
This will split an exe from it's params; stripping parenthesis from the exe; assumes clean data:
^(?:"([^"]+(?="))|([^\s]+))["]{0,1} +(.+)$
You will have two matches at a time, of three match groups:
The exe if it was wrapped in parenthesis
The exe if it was not wrapped in parenthesis
The clump of parameters
Examples:
"C:\WINDOWS\system32\cmd.exe" /c echo this
Match 1: C:\WINDOWS\system32\cmd.exe
Match 2: $null
Match 3: /c echo this
C:\WINDOWS\system32\cmd.exe /c echo this
Match 1: $null
Match 2: C:\WINDOWS\system32\cmd.exe
Match 3: /c echo this
"C:\Program Files\foo\bar.exe" /run
Match 1: C:\Program Files\foo\bar.exe
Match 2: $null
Match 3: /run
Thoughts:
I'm pretty sure that you would need to create a loop to capture a possibly infinite number of parameters.
This regex could easily be looped onto it's third match until the match fails; there are no more params.
If its just the quotes you are worried about, then just write a simple loop to dump character by character to a string ignoring the quotes.
Alternatively if you are using some string manipulation library, you can use it to remove all quotes and then concatenate them.
there's a python answer thus we shall have a ruby answer as well :)
require 'shellwords'
Shellwords.shellsplit '"param 1" param2 "param 3"'
#=> ["param 1", "param2", "param 3"] or :
'"param 1" param2 "param 3"'.shellsplit
Though answer is not RegEx specific but answers Python commandline arg parsing:
dash and double dash flags
int/float conversion based on SO answer
import sys
def parse_cmd_args():
_sys_args = sys.argv
_parts = {}
_key = "script"
_parts[_key] = [_sys_args.pop(0)]
for _part in _sys_args:
# Parse numeric values float and integers
if _part.replace("-", "1", 1).replace(".", "1").replace(",", "").isdigit():
_part = int(_part) if '.' not in _part and float(_part)/int(_part) == 1 else float(_part)
_parts[_key].append(_part)
elif "=" in _part:
_part = _part.split("=")
_parts[_part[0].strip("-")] = _part[1].strip().split(",")
elif _part.startswith(("-")):
_key = _part.strip("-")
_parts[_key] = []
else:
_parts[_key].extend(_part.split(","))
return _parts
Something like:
"(?:(?<=")([^"]+)"\s*)|\s*([^"\s]+)
or a simpler one:
"([^"]+)"|\s*([^"\s]+)
(just for the sake of finding a regexp ;) )
Apply it several time, and the group n°1 will give you the parameter, whether it is surrounded by double quotes or not.
If you are looking to parse the command and the parameters I use the following (with ^$ matching at line breaks aka multiline):
(?<cmd>^"[^"]*"|\S*) *(?<prm>.*)?
In case you want to use it in your C# code, here it is properly escaped:
try {
Regex RegexObj = new Regex("(?<cmd>^\\\"[^\\\"]*\\\"|\\S*) *(?<prm>.*)?");
} catch (ArgumentException ex) {
// Syntax error in the regular expression
}
It will parse the following and know what is the command versus the parameters:
"c:\program files\myapp\app.exe" p1 p2 "p3 with space"
app.exe p1 p2 "p3 with space"
app.exe
Here's a solution in Perl:
#!/usr/bin/perl
sub parse_arguments {
my $text = shift;
my $i = 0;
my #args;
while ($text ne '') {
$text =~ s{^\s*(['"]?)}{}; # look for (and remove) leading quote
my $delimiter = ($1 || ' '); # use space if not quoted
if ($text =~ s{^(([^$delimiter\\]|\\.|\\$)+)($delimiter|$)}{}) {
$args[$i++] = $1; # acquired an argument; save it
}
}
return #args;
}
my $line = <<'EOS';
"param 1" param\ 2 "pa\"ram' '3" 'pa\'ram" "4'
EOS
say "ARG: $_" for parse_arguments($line);
Output:
ARG: param 1
ARG: param\ 2
ARG: pa"ram' '3
ARG: pa'ram" "4
Note the following:
Arguments can be quoted with either " or ' (with the "other"
quote type treated as a regular character for that argument).
Spaces and quotes in arguments can be escaped with \.
The solution can be adapted to other languages. The basic approach is to (1) determine the delimiter character for the next string, (2) extract the next argument up to an unescaped occurrence of that delimiter or to the end-of-string, then (3) repeat until empty.
\s*("[^"]+"|[^\s"]+)
that's it
(reading your question again, just prior to posting I note you say command line LIKE string, thus this information may not be useful to you, but as I have written it I will post anyway - please disregard if I have missunderstood your question.)
If you clarify your question I will try to help but from the general comments you have made i would say dont do that :-), you are asking for a regexp to split a series of parmeters into an array. Instead of doing this yourself I would strongly suggest you consider using getopt, there are versions of this library for most programming languages. Getopt will do what you are asking and scales to manage much more sophisticated argument processing should you require that in the future.
If you let me know what language you are using I will try and post a sample for you.
Here are a sample of the home pages:
http://www.codeplex.com/getopt
(.NET)
http://www.urbanophile.com/arenn/hacking/download.html
(java)
A sample (from the java page above)
Getopt g = new Getopt("testprog", argv, "ab:c::d");
//
int c;
String arg;
while ((c = g.getopt()) != -1)
{
switch(c)
{
case 'a':
case 'd':
System.out.print("You picked " + (char)c + "\n");
break;
//
case 'b':
case 'c':
arg = g.getOptarg();
System.out.print("You picked " + (char)c +
" with an argument of " +
((arg != null) ? arg : "null") + "\n");
break;
//
case '?':
break; // getopt() already printed an error
//
default:
System.out.print("getopt() returned " + c + "\n");
}
}
Related
I need a regular expression to select all the text between two outer brackets.
Example:
START_TEXT(text here(possible text)text(possible text(more text)))END_TXT
^ ^
Result:
(text here(possible text)text(possible text(more text)))
I want to add this answer for quickreference. Feel free to update.
.NET Regex using balancing groups:
\((?>\((?<c>)|[^()]+|\)(?<-c>))*(?(c)(?!))\)
Where c is used as the depth counter.
Demo at Regexstorm.com
Stack Overflow: Using RegEx to balance match parenthesis
Wes' Puzzling Blog: Matching Balanced Constructs with .NET Regular Expressions
Greg Reinacker's Weblog: Nested Constructs in Regular Expressions
PCRE using a recursive pattern:
\((?:[^)(]+|(?R))*+\)
Demo at regex101; Or without alternation:
\((?:[^)(]*(?R)?)*+\)
Demo at regex101; Or unrolled for performance:
\([^)(]*+(?:(?R)[^)(]*)*+\)
Demo at regex101; The pattern is pasted at (?R) which represents (?0).
Perl, PHP, Notepad++, R: perl=TRUE, Python: PyPI regex module with (?V1) for Perl behaviour.
(the new version of PyPI regex package already defaults to this → DEFAULT_VERSION = VERSION1)
Ruby using subexpression calls:
With Ruby 2.0 \g<0> can be used to call full pattern.
\((?>[^)(]+|\g<0>)*\)
Demo at Rubular; Ruby 1.9 only supports capturing group recursion:
(\((?>[^)(]+|\g<1>)*\))
Demo at Rubular (atomic grouping since Ruby 1.9.3)
JavaScript API :: XRegExp.matchRecursive
XRegExp.matchRecursive(str, '\\(', '\\)', 'g');
Java: An interesting idea using forward references by #jaytea.
Without recursion up to 3 levels of nesting:
(JS, Java and other regex flavors)
To prevent runaway if unbalanced, with * on innermost [)(] only.
\((?:[^)(]|\((?:[^)(]|\((?:[^)(]|\([^)(]*\))*\))*\))*\)
Demo at regex101; Or unrolled for better performance (preferred).
\([^)(]*(?:\([^)(]*(?:\([^)(]*(?:\([^)(]*\)[^)(]*)*\)[^)(]*)*\)[^)(]*)*\)
Demo at regex101; Deeper nesting needs to be added as required.
Reference - What does this regex mean?
RexEgg.com - Recursive Regular Expressions
Regular-Expressions.info - Regular Expression Recursion
Mastering Regular Expressions - Jeffrey E.F. Friedl 1 2 3 4
Regular expressions are the wrong tool for the job because you are dealing with nested structures, i.e. recursion.
But there is a simple algorithm to do this, which I described in more detail in this answer to a previous question. The gist is to write code which scans through the string keeping a counter of the open parentheses which have not yet been matched by a closing parenthesis. When that counter returns to zero, then you know you've reached the final closing parenthesis.
You can use regex recursion:
\(([^()]|(?R))*\)
[^\(]*(\(.*\))[^\)]*
[^\(]* matches everything that isn't an opening bracket at the beginning of the string, (\(.*\)) captures the required substring enclosed in brackets, and [^\)]* matches everything that isn't a closing bracket at the end of the string. Note that this expression does not attempt to match brackets; a simple parser (see dehmann's answer) would be more suitable for that.
This answer explains the theoretical limitation of why regular expressions are not the right tool for this task.
Regular expressions can not do this.
Regular expressions are based on a computing model known as Finite State Automata (FSA). As the name indicates, a FSA can remember only the current state, it has no information about the previous states.
In the above diagram, S1 and S2 are two states where S1 is the starting and final step. So if we try with the string 0110 , the transition goes as follows:
0 1 1 0
-> S1 -> S2 -> S2 -> S2 ->S1
In the above steps, when we are at second S2 i.e. after parsing 01 of 0110, the FSA has no information about the previous 0 in 01 as it can only remember the current state and the next input symbol.
In the above problem, we need to know the no of opening parenthesis; this means it has to be stored at some place. But since FSAs can not do that, a regular expression can not be written.
However, an algorithm can be written to do this task. Algorithms are generally falls under Pushdown Automata (PDA). PDA is one level above of FSA. PDA has an additional stack to store some additional information. PDAs can be used to solve the above problem, because we can 'push' the opening parenthesis in the stack and 'pop' them once we encounter a closing parenthesis. If at the end, stack is empty, then opening parenthesis and closing parenthesis matches. Otherwise not.
(?<=\().*(?=\))
If you want to select text between two matching parentheses, you are out of luck with regular expressions. This is impossible(*).
This regex just returns the text between the first opening and the last closing parentheses in your string.
(*) Unless your regex engine has features like balancing groups or recursion. The number of engines that support such features is slowly growing, but they are still not a commonly available.
It is actually possible to do it using .NET regular expressions, but it is not trivial, so read carefully.
You can read a nice article here. You also may need to read up on .NET regular expressions. You can start reading here.
Angle brackets <> were used because they do not require escaping.
The regular expression looks like this:
<
[^<>]*
(
(
(?<Open><)
[^<>]*
)+
(
(?<Close-Open>>)
[^<>]*
)+
)*
(?(Open)(?!))
>
I was also stuck in this situation when dealing with nested patterns and regular-expressions is the right tool to solve such problems.
/(\((?>[^()]+|(?1))*\))/
This is the definitive regex:
\(
(?<arguments>
(
([^\(\)']*) |
(\([^\(\)']*\)) |
'(.*?)'
)*
)
\)
Example:
input: ( arg1, arg2, arg3, (arg4), '(pip' )
output: arg1, arg2, arg3, (arg4), '(pip'
note that the '(pip' is correctly managed as string.
(tried in regulator: http://sourceforge.net/projects/regulator/)
I have written a little JavaScript library called balanced to help with this task. You can accomplish this by doing
balanced.matches({
source: source,
open: '(',
close: ')'
});
You can even do replacements:
balanced.replacements({
source: source,
open: '(',
close: ')',
replace: function (source, head, tail) {
return head + source + tail;
}
});
Here's a more complex and interactive example JSFiddle.
Adding to bobble bubble's answer, there are other regex flavors where recursive constructs are supported.
Lua
Use %b() (%b{} / %b[] for curly braces / square brackets):
for s in string.gmatch("Extract (a(b)c) and ((d)f(g))", "%b()") do print(s) end (see demo)
Raku (former Perl6):
Non-overlapping multiple balanced parentheses matches:
my regex paren_any { '(' ~ ')' [ <-[()]>+ || <&paren_any> ]* }
say "Extract (a(b)c) and ((d)f(g))" ~~ m:g/<&paren_any>/;
# => (「(a(b)c)」 「((d)f(g))」)
Overlapping multiple balanced parentheses matches:
say "Extract (a(b)c) and ((d)f(g))" ~~ m:ov:g/<&paren_any>/;
# => (「(a(b)c)」 「(b)」 「((d)f(g))」 「(d)」 「(g)」)
See demo.
Python re non-regex solution
See poke's answer for How to get an expression between balanced parentheses.
Java customizable non-regex solution
Here is a customizable solution allowing single character literal delimiters in Java:
public static List<String> getBalancedSubstrings(String s, Character markStart,
Character markEnd, Boolean includeMarkers)
{
List<String> subTreeList = new ArrayList<String>();
int level = 0;
int lastOpenDelimiter = -1;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == markStart) {
level++;
if (level == 1) {
lastOpenDelimiter = (includeMarkers ? i : i + 1);
}
}
else if (c == markEnd) {
if (level == 1) {
subTreeList.add(s.substring(lastOpenDelimiter, (includeMarkers ? i + 1 : i)));
}
if (level > 0) level--;
}
}
return subTreeList;
}
}
Sample usage:
String s = "some text(text here(possible text)text(possible text(more text)))end text";
List<String> balanced = getBalancedSubstrings(s, '(', ')', true);
System.out.println("Balanced substrings:\n" + balanced);
// => [(text here(possible text)text(possible text(more text)))]
The regular expression using Ruby (version 1.9.3 or above):
/(?<match>\((?:\g<match>|[^()]++)*\))/
Demo on rubular
The answer depends on whether you need to match matching sets of brackets, or merely the first open to the last close in the input text.
If you need to match matching nested brackets, then you need something more than regular expressions. - see #dehmann
If it's just first open to last close see #Zach
Decide what you want to happen with:
abc ( 123 ( foobar ) def ) xyz ) ghij
You need to decide what your code needs to match in this case.
"""
Here is a simple python program showing how to use regular
expressions to write a paren-matching recursive parser.
This parser recognises items enclosed by parens, brackets,
braces and <> symbols, but is adaptable to any set of
open/close patterns. This is where the re package greatly
assists in parsing.
"""
import re
# The pattern below recognises a sequence consisting of:
# 1. Any characters not in the set of open/close strings.
# 2. One of the open/close strings.
# 3. The remainder of the string.
#
# There is no reason the opening pattern can't be the
# same as the closing pattern, so quoted strings can
# be included. However quotes are not ignored inside
# quotes. More logic is needed for that....
pat = re.compile("""
( .*? )
( \( | \) | \[ | \] | \{ | \} | \< | \> |
\' | \" | BEGIN | END | $ )
( .* )
""", re.X)
# The keys to the dictionary below are the opening strings,
# and the values are the corresponding closing strings.
# For example "(" is an opening string and ")" is its
# closing string.
matching = { "(" : ")",
"[" : "]",
"{" : "}",
"<" : ">",
'"' : '"',
"'" : "'",
"BEGIN" : "END" }
# The procedure below matches string s and returns a
# recursive list matching the nesting of the open/close
# patterns in s.
def matchnested(s, term=""):
lst = []
while True:
m = pat.match(s)
if m.group(1) != "":
lst.append(m.group(1))
if m.group(2) == term:
return lst, m.group(3)
if m.group(2) in matching:
item, s = matchnested(m.group(3), matching[m.group(2)])
lst.append(m.group(2))
lst.append(item)
lst.append(matching[m.group(2)])
else:
raise ValueError("After <<%s %s>> expected %s not %s" %
(lst, s, term, m.group(2)))
# Unit test.
if __name__ == "__main__":
for s in ("simple string",
""" "double quote" """,
""" 'single quote' """,
"one'two'three'four'five'six'seven",
"one(two(three(four)five)six)seven",
"one(two(three)four)five(six(seven)eight)nine",
"one(two)three[four]five{six}seven<eight>nine",
"one(two[three{four<five>six}seven]eight)nine",
"oneBEGINtwo(threeBEGINfourENDfive)sixENDseven",
"ERROR testing ((( mismatched ))] parens"):
print "\ninput", s
try:
lst, s = matchnested(s)
print "output", lst
except ValueError as e:
print str(e)
print "done"
You need the first and last parentheses. Use something like this:
str.indexOf('('); - it will give you first occurrence
str.lastIndexOf(')'); - last one
So you need a string between,
String searchedString = str.substring(str1.indexOf('('),str1.lastIndexOf(')');
because js regex doesn't support recursive match, i can't make balanced parentheses matching work.
so this is a simple javascript for loop version that make "method(arg)" string into array
push(number) map(test(a(a()))) bass(wow, abc)
$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)
const parser = str => {
let ops = []
let method, arg
let isMethod = true
let open = []
for (const char of str) {
// skip whitespace
if (char === ' ') continue
// append method or arg string
if (char !== '(' && char !== ')') {
if (isMethod) {
(method ? (method += char) : (method = char))
} else {
(arg ? (arg += char) : (arg = char))
}
}
if (char === '(') {
// nested parenthesis should be a part of arg
if (!isMethod) arg += char
isMethod = false
open.push(char)
} else if (char === ')') {
open.pop()
// check end of arg
if (open.length < 1) {
isMethod = true
ops.push({ method, arg })
method = arg = undefined
} else {
arg += char
}
}
}
return ops
}
// const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`)
const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`)
console.log(test)
the result is like
[ { method: 'push', arg: 'number' },
{ method: 'map', arg: 'test(a(a()))' },
{ method: 'bass', arg: 'wow,abc' } ]
[ { method: '$$', arg: 'groups' },
{ method: 'filter',
arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' },
{ method: 'pickBy', arg: '_id,type' },
{ method: 'map', arg: 'test()' },
{ method: 'as', arg: 'groups' } ]
While so many answers mention this in some form by saying that regex does not support recursive matching and so on, the primary reason for this lies in the roots of the Theory of Computation.
Language of the form {a^nb^n | n>=0} is not regular. Regex can only match things that form part of the regular set of languages.
Read more # here
I didn't use regex since it is difficult to deal with nested code. So this snippet should be able to allow you to grab sections of code with balanced brackets:
def extract_code(data):
""" returns an array of code snippets from a string (data)"""
start_pos = None
end_pos = None
count_open = 0
count_close = 0
code_snippets = []
for i,v in enumerate(data):
if v =='{':
count_open+=1
if not start_pos:
start_pos= i
if v=='}':
count_close +=1
if count_open == count_close and not end_pos:
end_pos = i+1
if start_pos and end_pos:
code_snippets.append((start_pos,end_pos))
start_pos = None
end_pos = None
return code_snippets
I used this to extract code snippets from a text file.
This do not fully address the OP question but I though it may be useful to some coming here to search for nested structure regexp:
Parse parmeters from function string (with nested structures) in javascript
Match structures like:
matches brackets, square brackets, parentheses, single and double quotes
Here you can see generated regexp in action
/**
* get param content of function string.
* only params string should be provided without parentheses
* WORK even if some/all params are not set
* #return [param1, param2, param3]
*/
exports.getParamsSAFE = (str, nbParams = 3) => {
const nextParamReg = /^\s*((?:(?:['"([{](?:[^'"()[\]{}]*?|['"([{](?:[^'"()[\]{}]*?|['"([{][^'"()[\]{}]*?['")}\]])*?['")}\]])*?['")}\]])|[^,])*?)\s*(?:,|$)/;
const params = [];
while (str.length) { // this is to avoid a BIG performance issue in javascript regexp engine
str = str.replace(nextParamReg, (full, p1) => {
params.push(p1);
return '';
});
}
return params;
};
This might help to match balanced parenthesis.
\s*\w+[(][^+]*[)]\s*
This one also worked
re.findall(r'\(.+\)', s)
I need a regular expression to select all the text between two outer brackets.
Example:
START_TEXT(text here(possible text)text(possible text(more text)))END_TXT
^ ^
Result:
(text here(possible text)text(possible text(more text)))
I want to add this answer for quickreference. Feel free to update.
.NET Regex using balancing groups:
\((?>\((?<c>)|[^()]+|\)(?<-c>))*(?(c)(?!))\)
Where c is used as the depth counter.
Demo at Regexstorm.com
Stack Overflow: Using RegEx to balance match parenthesis
Wes' Puzzling Blog: Matching Balanced Constructs with .NET Regular Expressions
Greg Reinacker's Weblog: Nested Constructs in Regular Expressions
PCRE using a recursive pattern:
\((?:[^)(]+|(?R))*+\)
Demo at regex101; Or without alternation:
\((?:[^)(]*(?R)?)*+\)
Demo at regex101; Or unrolled for performance:
\([^)(]*+(?:(?R)[^)(]*)*+\)
Demo at regex101; The pattern is pasted at (?R) which represents (?0).
Perl, PHP, Notepad++, R: perl=TRUE, Python: PyPI regex module with (?V1) for Perl behaviour.
(the new version of PyPI regex package already defaults to this → DEFAULT_VERSION = VERSION1)
Ruby using subexpression calls:
With Ruby 2.0 \g<0> can be used to call full pattern.
\((?>[^)(]+|\g<0>)*\)
Demo at Rubular; Ruby 1.9 only supports capturing group recursion:
(\((?>[^)(]+|\g<1>)*\))
Demo at Rubular (atomic grouping since Ruby 1.9.3)
JavaScript API :: XRegExp.matchRecursive
XRegExp.matchRecursive(str, '\\(', '\\)', 'g');
Java: An interesting idea using forward references by #jaytea.
Without recursion up to 3 levels of nesting:
(JS, Java and other regex flavors)
To prevent runaway if unbalanced, with * on innermost [)(] only.
\((?:[^)(]|\((?:[^)(]|\((?:[^)(]|\([^)(]*\))*\))*\))*\)
Demo at regex101; Or unrolled for better performance (preferred).
\([^)(]*(?:\([^)(]*(?:\([^)(]*(?:\([^)(]*\)[^)(]*)*\)[^)(]*)*\)[^)(]*)*\)
Demo at regex101; Deeper nesting needs to be added as required.
Reference - What does this regex mean?
RexEgg.com - Recursive Regular Expressions
Regular-Expressions.info - Regular Expression Recursion
Mastering Regular Expressions - Jeffrey E.F. Friedl 1 2 3 4
Regular expressions are the wrong tool for the job because you are dealing with nested structures, i.e. recursion.
But there is a simple algorithm to do this, which I described in more detail in this answer to a previous question. The gist is to write code which scans through the string keeping a counter of the open parentheses which have not yet been matched by a closing parenthesis. When that counter returns to zero, then you know you've reached the final closing parenthesis.
You can use regex recursion:
\(([^()]|(?R))*\)
[^\(]*(\(.*\))[^\)]*
[^\(]* matches everything that isn't an opening bracket at the beginning of the string, (\(.*\)) captures the required substring enclosed in brackets, and [^\)]* matches everything that isn't a closing bracket at the end of the string. Note that this expression does not attempt to match brackets; a simple parser (see dehmann's answer) would be more suitable for that.
This answer explains the theoretical limitation of why regular expressions are not the right tool for this task.
Regular expressions can not do this.
Regular expressions are based on a computing model known as Finite State Automata (FSA). As the name indicates, a FSA can remember only the current state, it has no information about the previous states.
In the above diagram, S1 and S2 are two states where S1 is the starting and final step. So if we try with the string 0110 , the transition goes as follows:
0 1 1 0
-> S1 -> S2 -> S2 -> S2 ->S1
In the above steps, when we are at second S2 i.e. after parsing 01 of 0110, the FSA has no information about the previous 0 in 01 as it can only remember the current state and the next input symbol.
In the above problem, we need to know the no of opening parenthesis; this means it has to be stored at some place. But since FSAs can not do that, a regular expression can not be written.
However, an algorithm can be written to do this task. Algorithms are generally falls under Pushdown Automata (PDA). PDA is one level above of FSA. PDA has an additional stack to store some additional information. PDAs can be used to solve the above problem, because we can 'push' the opening parenthesis in the stack and 'pop' them once we encounter a closing parenthesis. If at the end, stack is empty, then opening parenthesis and closing parenthesis matches. Otherwise not.
(?<=\().*(?=\))
If you want to select text between two matching parentheses, you are out of luck with regular expressions. This is impossible(*).
This regex just returns the text between the first opening and the last closing parentheses in your string.
(*) Unless your regex engine has features like balancing groups or recursion. The number of engines that support such features is slowly growing, but they are still not a commonly available.
It is actually possible to do it using .NET regular expressions, but it is not trivial, so read carefully.
You can read a nice article here. You also may need to read up on .NET regular expressions. You can start reading here.
Angle brackets <> were used because they do not require escaping.
The regular expression looks like this:
<
[^<>]*
(
(
(?<Open><)
[^<>]*
)+
(
(?<Close-Open>>)
[^<>]*
)+
)*
(?(Open)(?!))
>
I was also stuck in this situation when dealing with nested patterns and regular-expressions is the right tool to solve such problems.
/(\((?>[^()]+|(?1))*\))/
This is the definitive regex:
\(
(?<arguments>
(
([^\(\)']*) |
(\([^\(\)']*\)) |
'(.*?)'
)*
)
\)
Example:
input: ( arg1, arg2, arg3, (arg4), '(pip' )
output: arg1, arg2, arg3, (arg4), '(pip'
note that the '(pip' is correctly managed as string.
(tried in regulator: http://sourceforge.net/projects/regulator/)
I have written a little JavaScript library called balanced to help with this task. You can accomplish this by doing
balanced.matches({
source: source,
open: '(',
close: ')'
});
You can even do replacements:
balanced.replacements({
source: source,
open: '(',
close: ')',
replace: function (source, head, tail) {
return head + source + tail;
}
});
Here's a more complex and interactive example JSFiddle.
Adding to bobble bubble's answer, there are other regex flavors where recursive constructs are supported.
Lua
Use %b() (%b{} / %b[] for curly braces / square brackets):
for s in string.gmatch("Extract (a(b)c) and ((d)f(g))", "%b()") do print(s) end (see demo)
Raku (former Perl6):
Non-overlapping multiple balanced parentheses matches:
my regex paren_any { '(' ~ ')' [ <-[()]>+ || <&paren_any> ]* }
say "Extract (a(b)c) and ((d)f(g))" ~~ m:g/<&paren_any>/;
# => (「(a(b)c)」 「((d)f(g))」)
Overlapping multiple balanced parentheses matches:
say "Extract (a(b)c) and ((d)f(g))" ~~ m:ov:g/<&paren_any>/;
# => (「(a(b)c)」 「(b)」 「((d)f(g))」 「(d)」 「(g)」)
See demo.
Python re non-regex solution
See poke's answer for How to get an expression between balanced parentheses.
Java customizable non-regex solution
Here is a customizable solution allowing single character literal delimiters in Java:
public static List<String> getBalancedSubstrings(String s, Character markStart,
Character markEnd, Boolean includeMarkers)
{
List<String> subTreeList = new ArrayList<String>();
int level = 0;
int lastOpenDelimiter = -1;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == markStart) {
level++;
if (level == 1) {
lastOpenDelimiter = (includeMarkers ? i : i + 1);
}
}
else if (c == markEnd) {
if (level == 1) {
subTreeList.add(s.substring(lastOpenDelimiter, (includeMarkers ? i + 1 : i)));
}
if (level > 0) level--;
}
}
return subTreeList;
}
}
Sample usage:
String s = "some text(text here(possible text)text(possible text(more text)))end text";
List<String> balanced = getBalancedSubstrings(s, '(', ')', true);
System.out.println("Balanced substrings:\n" + balanced);
// => [(text here(possible text)text(possible text(more text)))]
The regular expression using Ruby (version 1.9.3 or above):
/(?<match>\((?:\g<match>|[^()]++)*\))/
Demo on rubular
The answer depends on whether you need to match matching sets of brackets, or merely the first open to the last close in the input text.
If you need to match matching nested brackets, then you need something more than regular expressions. - see #dehmann
If it's just first open to last close see #Zach
Decide what you want to happen with:
abc ( 123 ( foobar ) def ) xyz ) ghij
You need to decide what your code needs to match in this case.
"""
Here is a simple python program showing how to use regular
expressions to write a paren-matching recursive parser.
This parser recognises items enclosed by parens, brackets,
braces and <> symbols, but is adaptable to any set of
open/close patterns. This is where the re package greatly
assists in parsing.
"""
import re
# The pattern below recognises a sequence consisting of:
# 1. Any characters not in the set of open/close strings.
# 2. One of the open/close strings.
# 3. The remainder of the string.
#
# There is no reason the opening pattern can't be the
# same as the closing pattern, so quoted strings can
# be included. However quotes are not ignored inside
# quotes. More logic is needed for that....
pat = re.compile("""
( .*? )
( \( | \) | \[ | \] | \{ | \} | \< | \> |
\' | \" | BEGIN | END | $ )
( .* )
""", re.X)
# The keys to the dictionary below are the opening strings,
# and the values are the corresponding closing strings.
# For example "(" is an opening string and ")" is its
# closing string.
matching = { "(" : ")",
"[" : "]",
"{" : "}",
"<" : ">",
'"' : '"',
"'" : "'",
"BEGIN" : "END" }
# The procedure below matches string s and returns a
# recursive list matching the nesting of the open/close
# patterns in s.
def matchnested(s, term=""):
lst = []
while True:
m = pat.match(s)
if m.group(1) != "":
lst.append(m.group(1))
if m.group(2) == term:
return lst, m.group(3)
if m.group(2) in matching:
item, s = matchnested(m.group(3), matching[m.group(2)])
lst.append(m.group(2))
lst.append(item)
lst.append(matching[m.group(2)])
else:
raise ValueError("After <<%s %s>> expected %s not %s" %
(lst, s, term, m.group(2)))
# Unit test.
if __name__ == "__main__":
for s in ("simple string",
""" "double quote" """,
""" 'single quote' """,
"one'two'three'four'five'six'seven",
"one(two(three(four)five)six)seven",
"one(two(three)four)five(six(seven)eight)nine",
"one(two)three[four]five{six}seven<eight>nine",
"one(two[three{four<five>six}seven]eight)nine",
"oneBEGINtwo(threeBEGINfourENDfive)sixENDseven",
"ERROR testing ((( mismatched ))] parens"):
print "\ninput", s
try:
lst, s = matchnested(s)
print "output", lst
except ValueError as e:
print str(e)
print "done"
You need the first and last parentheses. Use something like this:
str.indexOf('('); - it will give you first occurrence
str.lastIndexOf(')'); - last one
So you need a string between,
String searchedString = str.substring(str1.indexOf('('),str1.lastIndexOf(')');
because js regex doesn't support recursive match, i can't make balanced parentheses matching work.
so this is a simple javascript for loop version that make "method(arg)" string into array
push(number) map(test(a(a()))) bass(wow, abc)
$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)
const parser = str => {
let ops = []
let method, arg
let isMethod = true
let open = []
for (const char of str) {
// skip whitespace
if (char === ' ') continue
// append method or arg string
if (char !== '(' && char !== ')') {
if (isMethod) {
(method ? (method += char) : (method = char))
} else {
(arg ? (arg += char) : (arg = char))
}
}
if (char === '(') {
// nested parenthesis should be a part of arg
if (!isMethod) arg += char
isMethod = false
open.push(char)
} else if (char === ')') {
open.pop()
// check end of arg
if (open.length < 1) {
isMethod = true
ops.push({ method, arg })
method = arg = undefined
} else {
arg += char
}
}
}
return ops
}
// const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`)
const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`)
console.log(test)
the result is like
[ { method: 'push', arg: 'number' },
{ method: 'map', arg: 'test(a(a()))' },
{ method: 'bass', arg: 'wow,abc' } ]
[ { method: '$$', arg: 'groups' },
{ method: 'filter',
arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' },
{ method: 'pickBy', arg: '_id,type' },
{ method: 'map', arg: 'test()' },
{ method: 'as', arg: 'groups' } ]
While so many answers mention this in some form by saying that regex does not support recursive matching and so on, the primary reason for this lies in the roots of the Theory of Computation.
Language of the form {a^nb^n | n>=0} is not regular. Regex can only match things that form part of the regular set of languages.
Read more # here
I didn't use regex since it is difficult to deal with nested code. So this snippet should be able to allow you to grab sections of code with balanced brackets:
def extract_code(data):
""" returns an array of code snippets from a string (data)"""
start_pos = None
end_pos = None
count_open = 0
count_close = 0
code_snippets = []
for i,v in enumerate(data):
if v =='{':
count_open+=1
if not start_pos:
start_pos= i
if v=='}':
count_close +=1
if count_open == count_close and not end_pos:
end_pos = i+1
if start_pos and end_pos:
code_snippets.append((start_pos,end_pos))
start_pos = None
end_pos = None
return code_snippets
I used this to extract code snippets from a text file.
This do not fully address the OP question but I though it may be useful to some coming here to search for nested structure regexp:
Parse parmeters from function string (with nested structures) in javascript
Match structures like:
matches brackets, square brackets, parentheses, single and double quotes
Here you can see generated regexp in action
/**
* get param content of function string.
* only params string should be provided without parentheses
* WORK even if some/all params are not set
* #return [param1, param2, param3]
*/
exports.getParamsSAFE = (str, nbParams = 3) => {
const nextParamReg = /^\s*((?:(?:['"([{](?:[^'"()[\]{}]*?|['"([{](?:[^'"()[\]{}]*?|['"([{][^'"()[\]{}]*?['")}\]])*?['")}\]])*?['")}\]])|[^,])*?)\s*(?:,|$)/;
const params = [];
while (str.length) { // this is to avoid a BIG performance issue in javascript regexp engine
str = str.replace(nextParamReg, (full, p1) => {
params.push(p1);
return '';
});
}
return params;
};
This might help to match balanced parenthesis.
\s*\w+[(][^+]*[)]\s*
This one also worked
re.findall(r'\(.+\)', s)
string MyName = " 'hi, load1', 'hi, load2', varthatnotstring ";
I want to use regex to split the above string at every ,, while preserving strings that are inside quotation.
As such, splitting MyName should yield:
1: 'hi, load1'
2: 'hi, load2'
3: varthatnotstring
I currently use regex MyR("(.),(.),(.*)");, but that gives me:
1: 'hi
2: load1'
3: 'hi
4: load2'
What regular-expression should I use?
Depending on how you want to handle certain corner cases, you can use the following:
std::regex reg(R"--((('.*?')|[^,])+)--");
Step, by step:
R"--(...)--" Is syntax for raw string literals, so we don't have to worry about escaping. We don't need it here, but I'm using them by default for regex strings.
('.*?') all characters between (and including) two apostrophes (non greedy)
[^,] anything that is not a comma
(('.*?')|[^,])+ arbitrary sequence of non-,-characters or '...'-sequences.
(Note: the ('.*?') part has to come first)
So this will also match e.g. tkasd 'rtzrze,123' as a single match. It will also NOT remove any whitespaces.
Usage:
std::regex reg(R"--((('.*?')|[^,])+)--");
std::string s = ",,t '123,4565',k ,'rt',t,z";
for (std::sregex_iterator rit(s.begin(), s.end(), reg), end{}; rit != end; ++rit) {
std::cout << rit->str() << std::endl;
}
Output:
t '123,4565'
k
'rt'
t
z
Edit:
I rarely use regular expressions, so any comments about possible improvements or gotchas are welcome. Maybe there is also an even better solution using regex_token_iterator.
I need to determine whether a string begins with a number - I've tried the following to no avail:
if (matches("^[0-9].*)", upper(text))) str = "Title"""
I'm new to DXL and Regex - what am I doing wrong?
You need the caret character to indicate a match only at the start of a string. I added the plus character to match all the numbers, although you might not need it for your situation. If you're only looking for numbers at the start, and don't care if there is anything following, you don't need anymore.
string str1 = "123abc"
string str2 = "abc123"
string strgx = "^[0-9]+"
Regexp rgx = regexp2(strgx)
if(rgx(str1)) { print str1[match 0] "\n" } else { print "no match\n" }
if(rgx(str2)) { print str2[match 0] "\n" } else { print "no match\n" }
The code block above will print:
123
no match
#mrhobo is correct, you want something like this:
Regexp numReg = "^[0-9]"
if(numReg text) str = "Title"
You don't need upper since you are just looking for numbers. Also matches is more for finding the part of the string that matches the expression. If you just want to check that the string as a whole matches the expression then the code above would be more efficient.
Good luck!
At least from example I found this example should work:
Regexp plural = regexp "^([0-9].*)$"
if plural "15systems" then print "yes"
Resource:
http://www.scenarioplus.org.uk/papers/dxl_regexp/dxl_regexp.htm
I need a regular expression to select all the text between two outer brackets.
Example:
START_TEXT(text here(possible text)text(possible text(more text)))END_TXT
^ ^
Result:
(text here(possible text)text(possible text(more text)))
I want to add this answer for quickreference. Feel free to update.
.NET Regex using balancing groups:
\((?>\((?<c>)|[^()]+|\)(?<-c>))*(?(c)(?!))\)
Where c is used as the depth counter.
Demo at Regexstorm.com
Stack Overflow: Using RegEx to balance match parenthesis
Wes' Puzzling Blog: Matching Balanced Constructs with .NET Regular Expressions
Greg Reinacker's Weblog: Nested Constructs in Regular Expressions
PCRE using a recursive pattern:
\((?:[^)(]+|(?R))*+\)
Demo at regex101; Or without alternation:
\((?:[^)(]*(?R)?)*+\)
Demo at regex101; Or unrolled for performance:
\([^)(]*+(?:(?R)[^)(]*)*+\)
Demo at regex101; The pattern is pasted at (?R) which represents (?0).
Perl, PHP, Notepad++, R: perl=TRUE, Python: PyPI regex module with (?V1) for Perl behaviour.
(the new version of PyPI regex package already defaults to this → DEFAULT_VERSION = VERSION1)
Ruby using subexpression calls:
With Ruby 2.0 \g<0> can be used to call full pattern.
\((?>[^)(]+|\g<0>)*\)
Demo at Rubular; Ruby 1.9 only supports capturing group recursion:
(\((?>[^)(]+|\g<1>)*\))
Demo at Rubular (atomic grouping since Ruby 1.9.3)
JavaScript API :: XRegExp.matchRecursive
XRegExp.matchRecursive(str, '\\(', '\\)', 'g');
Java: An interesting idea using forward references by #jaytea.
Without recursion up to 3 levels of nesting:
(JS, Java and other regex flavors)
To prevent runaway if unbalanced, with * on innermost [)(] only.
\((?:[^)(]|\((?:[^)(]|\((?:[^)(]|\([^)(]*\))*\))*\))*\)
Demo at regex101; Or unrolled for better performance (preferred).
\([^)(]*(?:\([^)(]*(?:\([^)(]*(?:\([^)(]*\)[^)(]*)*\)[^)(]*)*\)[^)(]*)*\)
Demo at regex101; Deeper nesting needs to be added as required.
Reference - What does this regex mean?
RexEgg.com - Recursive Regular Expressions
Regular-Expressions.info - Regular Expression Recursion
Mastering Regular Expressions - Jeffrey E.F. Friedl 1 2 3 4
Regular expressions are the wrong tool for the job because you are dealing with nested structures, i.e. recursion.
But there is a simple algorithm to do this, which I described in more detail in this answer to a previous question. The gist is to write code which scans through the string keeping a counter of the open parentheses which have not yet been matched by a closing parenthesis. When that counter returns to zero, then you know you've reached the final closing parenthesis.
You can use regex recursion:
\(([^()]|(?R))*\)
[^\(]*(\(.*\))[^\)]*
[^\(]* matches everything that isn't an opening bracket at the beginning of the string, (\(.*\)) captures the required substring enclosed in brackets, and [^\)]* matches everything that isn't a closing bracket at the end of the string. Note that this expression does not attempt to match brackets; a simple parser (see dehmann's answer) would be more suitable for that.
This answer explains the theoretical limitation of why regular expressions are not the right tool for this task.
Regular expressions can not do this.
Regular expressions are based on a computing model known as Finite State Automata (FSA). As the name indicates, a FSA can remember only the current state, it has no information about the previous states.
In the above diagram, S1 and S2 are two states where S1 is the starting and final step. So if we try with the string 0110 , the transition goes as follows:
0 1 1 0
-> S1 -> S2 -> S2 -> S2 ->S1
In the above steps, when we are at second S2 i.e. after parsing 01 of 0110, the FSA has no information about the previous 0 in 01 as it can only remember the current state and the next input symbol.
In the above problem, we need to know the no of opening parenthesis; this means it has to be stored at some place. But since FSAs can not do that, a regular expression can not be written.
However, an algorithm can be written to do this task. Algorithms are generally falls under Pushdown Automata (PDA). PDA is one level above of FSA. PDA has an additional stack to store some additional information. PDAs can be used to solve the above problem, because we can 'push' the opening parenthesis in the stack and 'pop' them once we encounter a closing parenthesis. If at the end, stack is empty, then opening parenthesis and closing parenthesis matches. Otherwise not.
(?<=\().*(?=\))
If you want to select text between two matching parentheses, you are out of luck with regular expressions. This is impossible(*).
This regex just returns the text between the first opening and the last closing parentheses in your string.
(*) Unless your regex engine has features like balancing groups or recursion. The number of engines that support such features is slowly growing, but they are still not a commonly available.
It is actually possible to do it using .NET regular expressions, but it is not trivial, so read carefully.
You can read a nice article here. You also may need to read up on .NET regular expressions. You can start reading here.
Angle brackets <> were used because they do not require escaping.
The regular expression looks like this:
<
[^<>]*
(
(
(?<Open><)
[^<>]*
)+
(
(?<Close-Open>>)
[^<>]*
)+
)*
(?(Open)(?!))
>
I was also stuck in this situation when dealing with nested patterns and regular-expressions is the right tool to solve such problems.
/(\((?>[^()]+|(?1))*\))/
This is the definitive regex:
\(
(?<arguments>
(
([^\(\)']*) |
(\([^\(\)']*\)) |
'(.*?)'
)*
)
\)
Example:
input: ( arg1, arg2, arg3, (arg4), '(pip' )
output: arg1, arg2, arg3, (arg4), '(pip'
note that the '(pip' is correctly managed as string.
(tried in regulator: http://sourceforge.net/projects/regulator/)
I have written a little JavaScript library called balanced to help with this task. You can accomplish this by doing
balanced.matches({
source: source,
open: '(',
close: ')'
});
You can even do replacements:
balanced.replacements({
source: source,
open: '(',
close: ')',
replace: function (source, head, tail) {
return head + source + tail;
}
});
Here's a more complex and interactive example JSFiddle.
Adding to bobble bubble's answer, there are other regex flavors where recursive constructs are supported.
Lua
Use %b() (%b{} / %b[] for curly braces / square brackets):
for s in string.gmatch("Extract (a(b)c) and ((d)f(g))", "%b()") do print(s) end (see demo)
Raku (former Perl6):
Non-overlapping multiple balanced parentheses matches:
my regex paren_any { '(' ~ ')' [ <-[()]>+ || <&paren_any> ]* }
say "Extract (a(b)c) and ((d)f(g))" ~~ m:g/<&paren_any>/;
# => (「(a(b)c)」 「((d)f(g))」)
Overlapping multiple balanced parentheses matches:
say "Extract (a(b)c) and ((d)f(g))" ~~ m:ov:g/<&paren_any>/;
# => (「(a(b)c)」 「(b)」 「((d)f(g))」 「(d)」 「(g)」)
See demo.
Python re non-regex solution
See poke's answer for How to get an expression between balanced parentheses.
Java customizable non-regex solution
Here is a customizable solution allowing single character literal delimiters in Java:
public static List<String> getBalancedSubstrings(String s, Character markStart,
Character markEnd, Boolean includeMarkers)
{
List<String> subTreeList = new ArrayList<String>();
int level = 0;
int lastOpenDelimiter = -1;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == markStart) {
level++;
if (level == 1) {
lastOpenDelimiter = (includeMarkers ? i : i + 1);
}
}
else if (c == markEnd) {
if (level == 1) {
subTreeList.add(s.substring(lastOpenDelimiter, (includeMarkers ? i + 1 : i)));
}
if (level > 0) level--;
}
}
return subTreeList;
}
}
Sample usage:
String s = "some text(text here(possible text)text(possible text(more text)))end text";
List<String> balanced = getBalancedSubstrings(s, '(', ')', true);
System.out.println("Balanced substrings:\n" + balanced);
// => [(text here(possible text)text(possible text(more text)))]
The regular expression using Ruby (version 1.9.3 or above):
/(?<match>\((?:\g<match>|[^()]++)*\))/
Demo on rubular
The answer depends on whether you need to match matching sets of brackets, or merely the first open to the last close in the input text.
If you need to match matching nested brackets, then you need something more than regular expressions. - see #dehmann
If it's just first open to last close see #Zach
Decide what you want to happen with:
abc ( 123 ( foobar ) def ) xyz ) ghij
You need to decide what your code needs to match in this case.
"""
Here is a simple python program showing how to use regular
expressions to write a paren-matching recursive parser.
This parser recognises items enclosed by parens, brackets,
braces and <> symbols, but is adaptable to any set of
open/close patterns. This is where the re package greatly
assists in parsing.
"""
import re
# The pattern below recognises a sequence consisting of:
# 1. Any characters not in the set of open/close strings.
# 2. One of the open/close strings.
# 3. The remainder of the string.
#
# There is no reason the opening pattern can't be the
# same as the closing pattern, so quoted strings can
# be included. However quotes are not ignored inside
# quotes. More logic is needed for that....
pat = re.compile("""
( .*? )
( \( | \) | \[ | \] | \{ | \} | \< | \> |
\' | \" | BEGIN | END | $ )
( .* )
""", re.X)
# The keys to the dictionary below are the opening strings,
# and the values are the corresponding closing strings.
# For example "(" is an opening string and ")" is its
# closing string.
matching = { "(" : ")",
"[" : "]",
"{" : "}",
"<" : ">",
'"' : '"',
"'" : "'",
"BEGIN" : "END" }
# The procedure below matches string s and returns a
# recursive list matching the nesting of the open/close
# patterns in s.
def matchnested(s, term=""):
lst = []
while True:
m = pat.match(s)
if m.group(1) != "":
lst.append(m.group(1))
if m.group(2) == term:
return lst, m.group(3)
if m.group(2) in matching:
item, s = matchnested(m.group(3), matching[m.group(2)])
lst.append(m.group(2))
lst.append(item)
lst.append(matching[m.group(2)])
else:
raise ValueError("After <<%s %s>> expected %s not %s" %
(lst, s, term, m.group(2)))
# Unit test.
if __name__ == "__main__":
for s in ("simple string",
""" "double quote" """,
""" 'single quote' """,
"one'two'three'four'five'six'seven",
"one(two(three(four)five)six)seven",
"one(two(three)four)five(six(seven)eight)nine",
"one(two)three[four]five{six}seven<eight>nine",
"one(two[three{four<five>six}seven]eight)nine",
"oneBEGINtwo(threeBEGINfourENDfive)sixENDseven",
"ERROR testing ((( mismatched ))] parens"):
print "\ninput", s
try:
lst, s = matchnested(s)
print "output", lst
except ValueError as e:
print str(e)
print "done"
You need the first and last parentheses. Use something like this:
str.indexOf('('); - it will give you first occurrence
str.lastIndexOf(')'); - last one
So you need a string between,
String searchedString = str.substring(str1.indexOf('('),str1.lastIndexOf(')');
because js regex doesn't support recursive match, i can't make balanced parentheses matching work.
so this is a simple javascript for loop version that make "method(arg)" string into array
push(number) map(test(a(a()))) bass(wow, abc)
$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)
const parser = str => {
let ops = []
let method, arg
let isMethod = true
let open = []
for (const char of str) {
// skip whitespace
if (char === ' ') continue
// append method or arg string
if (char !== '(' && char !== ')') {
if (isMethod) {
(method ? (method += char) : (method = char))
} else {
(arg ? (arg += char) : (arg = char))
}
}
if (char === '(') {
// nested parenthesis should be a part of arg
if (!isMethod) arg += char
isMethod = false
open.push(char)
} else if (char === ')') {
open.pop()
// check end of arg
if (open.length < 1) {
isMethod = true
ops.push({ method, arg })
method = arg = undefined
} else {
arg += char
}
}
}
return ops
}
// const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`)
const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`)
console.log(test)
the result is like
[ { method: 'push', arg: 'number' },
{ method: 'map', arg: 'test(a(a()))' },
{ method: 'bass', arg: 'wow,abc' } ]
[ { method: '$$', arg: 'groups' },
{ method: 'filter',
arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' },
{ method: 'pickBy', arg: '_id,type' },
{ method: 'map', arg: 'test()' },
{ method: 'as', arg: 'groups' } ]
While so many answers mention this in some form by saying that regex does not support recursive matching and so on, the primary reason for this lies in the roots of the Theory of Computation.
Language of the form {a^nb^n | n>=0} is not regular. Regex can only match things that form part of the regular set of languages.
Read more # here
I didn't use regex since it is difficult to deal with nested code. So this snippet should be able to allow you to grab sections of code with balanced brackets:
def extract_code(data):
""" returns an array of code snippets from a string (data)"""
start_pos = None
end_pos = None
count_open = 0
count_close = 0
code_snippets = []
for i,v in enumerate(data):
if v =='{':
count_open+=1
if not start_pos:
start_pos= i
if v=='}':
count_close +=1
if count_open == count_close and not end_pos:
end_pos = i+1
if start_pos and end_pos:
code_snippets.append((start_pos,end_pos))
start_pos = None
end_pos = None
return code_snippets
I used this to extract code snippets from a text file.
This do not fully address the OP question but I though it may be useful to some coming here to search for nested structure regexp:
Parse parmeters from function string (with nested structures) in javascript
Match structures like:
matches brackets, square brackets, parentheses, single and double quotes
Here you can see generated regexp in action
/**
* get param content of function string.
* only params string should be provided without parentheses
* WORK even if some/all params are not set
* #return [param1, param2, param3]
*/
exports.getParamsSAFE = (str, nbParams = 3) => {
const nextParamReg = /^\s*((?:(?:['"([{](?:[^'"()[\]{}]*?|['"([{](?:[^'"()[\]{}]*?|['"([{][^'"()[\]{}]*?['")}\]])*?['")}\]])*?['")}\]])|[^,])*?)\s*(?:,|$)/;
const params = [];
while (str.length) { // this is to avoid a BIG performance issue in javascript regexp engine
str = str.replace(nextParamReg, (full, p1) => {
params.push(p1);
return '';
});
}
return params;
};
This might help to match balanced parenthesis.
\s*\w+[(][^+]*[)]\s*
This one also worked
re.findall(r'\(.+\)', s)