Regex to search two string inside a method - regex

How to create a regex , so as to search two strings "greet" AND inside this method string "name" .
I tried
(^.*greet(\n|.|\t)*)(.*name*)
def greet(name):
print("Hello, " + name + ". Good morning!") <--- this name should be selected
def meet(name):
print("Lets meet, " + name )

I would use this regex:
greet([^\n]|\n+[^\S\n])*name
Here the strings greet and name are separated by characters that are not a linebreak ([^\n]) or, in the case, they must be eventually followed by a space that is not a linebreak ([^\S\n]). In this way you ensure that name is in the same method of greet.
See demo.

You can capture in a group what is between the parenthesis, and use a backreference \1 in the next line to match the same.
If you want to select it, you could also capture that in a group itself.
\bdef greet\(([^\s()]+)\):\r?\n.*(\1)
Regex demo
If it should be name only
\bdef greet\([^\s()]+\):\r?\n.*\b(name)\b
Regex demo

Related

How can I search and replace guids in Sublime 3

I have a textfile where I would like to replace all GUIDs with space.
I want:
92094, "970d6c9e-c199-40e3-80ea-14daf1141904"
91995, "970d6c9e-c199-40e3-80ea-14daf1141904"
87445, "f17e66ef-b1df-4270-8285-b3c15da366f7"
87298, "f17e66ef-b1df-4270-8285-b3c15da366f7"
96713, "3c28e493-015b-4b48-957f-fe3e7acc8412"
96759, "3c28e493-015b-4b48-957f-fe3e7acc8412"
94665, "87ac12a3-62ed-4e1d-a1a6-51ae05e01b1a"
94405, "87ac12a3-62ed-4e1d-a1a6-51ae05e01b1a"
To become:
92094,
91995,
87445,
87298,
96713,
96759,
94665,
94405,
How can i accomplish this in Sublime 3?
Ctrl+H
Find: "[\da-f-]{36}"
Replace: LEAVE EMPTY
Enable regex mode
Replace all
Explanation:
" : double quote
[ : start class character
\d : any digit
a-f : or letter from a to f
- : or a dash
]{36} : end class, 36 characters must be present
" : double quote
Result for given example:
92094,
91995,
87445,
87298,
96713,
96759,
94665,
94405,
Try doing a search for this pattern in regex search mode:
"[0-9a-z]{8}-[0-9a-z]{4}-[0-9a-z]{4}-[0-9a-z]{4}-[0-9a-z]{12}"
And then just replace with empty string. This should strip off the GUID, leaving you with the output you want.
Demo
Another regex solution involving a slightly different search-replace strategy where we don't care about the GUI format and simply get the first column:
Search for ([^,]*,).* (again don't forget to activate the regex mode .*).
Replace with $1.
Details about the regular expression
The idea here is to capture all first columns. A column here is defined by a sequence of
"some non-comma character": [^,]*
followed by a comma: [^,]*,
The first column can then be followed by anything .* (the GUI format doesn't matter): [^,]*,.*
Finally we need to capture the 1st column using group capturing: ([^,]*,).*
In the replace field we use a backreference $x which refers the the x-th capturing group.

Surrounding one group with special characters in using substitute in vim

Given string:
some_function(inputId = "select_something"),
(...)
some_other_function(inputId = "some_other_label")
I would like to arrive at:
some_function(inputId = ns("select_something")),
(...)
some_other_function(inputId = ns("some_other_label"))
The key change here is the element ns( ... ) that surrounds the string available in the "" after the inputId
Regex
So far, I have came up with this regex:
:%substitute/\(inputId\s=\s\)\(\"[a-zA-Z]"\)/\1ns(/2/cgI
However, when deployed, it produces an error:
E488: Trailing characters
A simpler version of that regex works, the syntax:
:%substitute/\(inputId\s=\s\)/\1ns(/cgI
would correctly inser ns( after finding inputId = and create string
some_other_function(inputId = ns("some_other_label")
Challenge
I'm struggling to match the remaining part of the string, ex. "select_something") and return it as:
"select_something")).
You have many problems with your regex.
[a-zA-Z] will only match one letter. Presumably you want to match everything up to the next ", so you'll need a \+ and you'll also need to match underscores too. I would recommend \w\+. Unless more than [a-zA-Z_] might be in the string, in which case I would do .\{-}.
You have a /2 instead of \2. This is why you're getting E488.
I would do this:
:%s/\(inputId = \)\(".\{-}\)"/\1ns(\2)/cgI
Or use the start match atom: (that is, \zs)
:%s/inputId = \zs\".\{-}"/ns(&)/cgI
You can use a negated character class "[^"]*" to match a quoted string:
%s/\(inputId\s*=\s*\)\("[^"]*"\)/\1ns(\2)/g

Regex to grab formulas

I am trying to parse a file that contains parameter attributes. The attributes are setup like this:
w=(nf*40e-9)*ng
but also like this:
par_nf=(1) * (ng)
The issue is, all of these parameter definitions are on a single line in the source file, and they are separated by spaces. So you might have a situation like this:
pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0
The current algorithm just splits the line on spaces and then for each token, the name is extracted from the LHS of the = and the value from the RHS. My thought is if I can create a Regex match based on spaces within parameter declarations, I can then remove just those spaces before feeding the line to the splitter/parser. I am having a tough time coming up with the appropriate Regex, however. Is it possible to create a regex that matches only spaces within parameter declarations, but ignores the spaces between parameter declarations?
Try this RegEx:
(?<=^|\s) # Start of each formula (start of line OR [space])
(?:.*?) # Attribute Name
= # =
(?: # Formula
(?!\s\w+=) # DO NOT Match [space] Word Characters = (Attr. Name)
[^=] # Any Character except =
)* # Formula Characters repeated any number of times
When checking formula characters, it uses a negative lookahead to check for a Space, followed by Word Characters (Attribute Name) and an =. If this is found, it will stop the match. The fact that the negative lookahead checks for a space means that it will stop without a trailing space at the end of the formula.
Live Demo on Regex101
Thanks to #Andy for the tip:
In this case I'll probably just match on the parameter name and equals, but replace the preceding whitespace with some other "parse-able" character to split on, like so:
(\s*)\w+[a-zA-Z_]=
Now my first capturing group can be used to insert something like a colon, semicolon, or line-break.
You need to add Perl tag. :-( Maybe this will help:
I ended up using this in C#. The idea was to break it into name value pairs, using a negative lookahead specified as the key to stop a match and start a new one. If this helps
var data = #"pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0";
var pattern = #"
(?<Key>[a-zA-Z_\s\d]+) # Key is any alpha, digit and _
= # = is a hard anchor
(?<Value>[.*+\-\\\/()\w\s]+) # Value is any combinations of text with space(s)
(\s|$) # Soft anchor of either a \s or EOB
((?!\s[a-zA-Z_\d\s]+\=)|$) # Negative lookahead to stop matching if a space then key then equal found or EOB
";
Regex.Matches(data, pattern, RegexOptions.IgnorePatternWhitespace | RegexOptions.ExplicitCapture)
.OfType<Match>()
.Select(mt => new
{
LHS = mt.Groups["Key"].Value,
RHS = mt.Groups["Value"].Value
});
Results:

How to better this regex?

I have a list of strings like this:
/soccer/poland/ekstraklasa-2008-2009/results/
/soccer/poland/orange-ekstraklasa-2007-2008/results/
/soccer/poland/orange-ekstraklasa-youth-2010-2011/results/
From each string I want to take a middle part resulting in respectively:
ekstraklasa
orange ekstraklasa
orange ekstraklasa youth
My code here does the job but it feels like it can be done in fewer steps and probably with regex alone.
name = re.search('/([-a-z\d]+)/results/', string).group(1) # take the middle part
name = re.search('[-a-z]+', name).group() # trim numbers
if name.endswith('-'):
name = name[:-1] # trim tailing `-` if needed
name = name.replace('-', ' ')
Can anyone see how make it better?
This regex should do the work:
/(?:\/\w+){2}\/([\w\-]+)(?:-\d+){2}/
Explanation:
(?:\/\w+){2} - eat the first two words delimited by /
\/ - eat the next /
([\w\-]+)- match the word characters of hyphens (this is what we're looking for)
(?:-\d+){2} - eat the hyphens and the numbers after the part we're looking for
The result is in the first match group
I cant test it because i am not using python, but i would use an Expression like
^(/soccer/poland/)([a-z\-]*)(.*)$
or
^(/[a-z]*/[a-z]*/)([a-z\-]*)(.*)$
This Expressen works like "/soccer/poland/" at the beginning, than "everything with a to z (small) or -" and the rest of the string.
And than taking 2nd Group!
The Groups should hold this Strings:
/soccer/poland/
orange-ekstraklasa-youth-
2010-2011/results/
And then simply replacing "-" with " " and after that TRIM Spaces.
PS: If ur Using regex101.com e.g., u need to escape / AND just use one Row of String!
Expression
^(\/soccer\/poland\/)([a-z\-]*)(.*)$
And one Row of ur String.
/soccer/poland/orange-ekstraklasa-youth-2010-2011/results/
If u prefere to use the Expression not just for soccer and poland, use
^(\/[a-z]*\/[a-z]*\/)([a-z\-]*)(.*)$

Regex for quoted string with escaping quotes

How do I get the substring " It's big \"problem " using a regular expression?
s = ' function(){ return " It\'s big \"problem "; }';
/"(?:[^"\\]|\\.)*"/
Works in The Regex Coach and PCRE Workbench.
Example of test in JavaScript:
var s = ' function(){ return " Is big \\"problem\\", \\no? "; }';
var m = s.match(/"(?:[^"\\]|\\.)*"/);
if (m != null)
alert(m);
This one comes from nanorc.sample available in many linux distros. It is used for syntax highlighting of C style strings
\"(\\.|[^\"])*\"
As provided by ePharaoh, the answer is
/"([^"\\]*(\\.[^"\\]*)*)"/
To have the above apply to either single quoted or double quoted strings, use
/"([^"\\]*(\\.[^"\\]*)*)"|\'([^\'\\]*(\\.[^\'\\]*)*)\'/
Most of the solutions provided here use alternative repetition paths i.e. (A|B)*.
You may encounter stack overflows on large inputs since some pattern compiler implements this using recursion.
Java for instance: http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6337993
Something like this:
"(?:[^"\\]*(?:\\.)?)*", or the one provided by Guy Bedford will reduce the amount of parsing steps avoiding most stack overflows.
/(["\']).*?(?<!\\)(\\\\)*\1/is
should work with any quoted string
"(?:\\"|.)*?"
Alternating the \" and the . passes over escaped quotes while the lazy quantifier *? ensures that you don't go past the end of the quoted string. Works with .NET Framework RE classes
/"(?:[^"\\]++|\\.)*+"/
Taken straight from man perlre on a Linux system with Perl 5.22.0 installed.
As an optimization, this regex uses the 'posessive' form of both + and * to prevent backtracking, for it is known beforehand that a string without a closing quote wouldn't match in any case.
This one works perfect on PCRE and does not fall with StackOverflow.
"(.*?[^\\])??((\\\\)+)?+"
Explanation:
Every quoted string starts with Char: " ;
It may contain any number of any characters: .*? {Lazy match}; ending with non escape character [^\\];
Statement (2) is Lazy(!) optional because string can be empty(""). So: (.*?[^\\])??
Finally, every quoted string ends with Char("), but it can be preceded with even number of escape sign pairs (\\\\)+; and it is Greedy(!) optional: ((\\\\)+)?+ {Greedy matching}, bacause string can be empty or without ending pairs!
An option that has not been touched on before is:
Reverse the string.
Perform the matching on the reversed string.
Re-reverse the matched strings.
This has the added bonus of being able to correctly match escaped open tags.
Lets say you had the following string; String \"this "should" NOT match\" and "this \"should\" match"
Here, \"this "should" NOT match\" should not be matched and "should" should be.
On top of that this \"should\" match should be matched and \"should\" should not.
First an example.
// The input string.
const myString = 'String \\"this "should" NOT match\\" and "this \\"should\\" match"';
// The RegExp.
const regExp = new RegExp(
// Match close
'([\'"])(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))' +
'((?:' +
// Match escaped close quote
'(?:\\1(?=(?:[\\\\]{2})*[\\\\](?![\\\\])))|' +
// Match everything thats not the close quote
'(?:(?!\\1).)' +
'){0,})' +
// Match open
'(\\1)(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))',
'g'
);
// Reverse the matched strings.
matches = myString
// Reverse the string.
.split('').reverse().join('')
// '"hctam "\dluohs"\ siht" dna "\hctam TON "dluohs" siht"\ gnirtS'
// Match the quoted
.match(regExp)
// ['"hctam "\dluohs"\ siht"', '"dluohs"']
// Reverse the matches
.map(x => x.split('').reverse().join(''))
// ['"this \"should\" match"', '"should"']
// Re order the matches
.reverse();
// ['"should"', '"this \"should\" match"']
Okay, now to explain the RegExp.
This is the regexp can be easily broken into three pieces. As follows:
# Part 1
(['"]) # Match a closing quotation mark " or '
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
# Part 2
((?: # Match inside the quotes
(?: # Match option 1:
\1 # Match the closing quote
(?= # As long as it's followed by
(?:\\\\)* # A pair of escape characters
\\ #
(?![\\]) # As long as that's not followed by an escape
) # and a single escape
)| # OR
(?: # Match option 2:
(?!\1). # Any character that isn't the closing quote
)
)*) # Match the group 0 or more times
# Part 3
(\1) # Match an open quotation mark that is the same as the closing one
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
This is probably a lot clearer in image form: generated using Jex's Regulex
Image on github (JavaScript Regular Expression Visualizer.)
Sorry, I don't have a high enough reputation to include images, so, it's just a link for now.
Here is a gist of an example function using this concept that's a little more advanced: https://gist.github.com/scagood/bd99371c072d49a4fee29d193252f5fc#file-matchquotes-js
here is one that work with both " and ' and you easily add others at the start.
("|')(?:\\\1|[^\1])*?\1
it uses the backreference (\1) match exactley what is in the first group (" or ').
http://www.regular-expressions.info/backref.html
One has to remember that regexps aren't a silver bullet for everything string-y. Some stuff are simpler to do with a cursor and linear, manual, seeking. A CFL would do the trick pretty trivially, but there aren't many CFL implementations (afaik).
A more extensive version of https://stackoverflow.com/a/10786066/1794894
/"([^"\\]{50,}(\\.[^"\\]*)*)"|\'[^\'\\]{50,}(\\.[^\'\\]*)*\'|“[^”\\]{50,}(\\.[^“\\]*)*”/
This version also contains
Minimum quote length of 50
Extra type of quotes (open “ and close ”)
If it is searched from the beginning, maybe this can work?
\"((\\\")|[^\\])*\"
I faced a similar problem trying to remove quoted strings that may interfere with parsing of some files.
I ended up with a two-step solution that beats any convoluted regex you can come up with:
line = line.replace("\\\"","\'"); // Replace escaped quotes with something easier to handle
line = line.replaceAll("\"([^\"]*)\"","\"x\""); // Simple is beautiful
Easier to read and probably more efficient.
If your IDE is IntelliJ Idea, you can forget all these headaches and store your regex into a String variable and as you copy-paste it inside the double-quote it will automatically change to a regex acceptable format.
example in Java:
String s = "\"en_usa\":[^\\,\\}]+";
now you can use this variable in your regexp or anywhere.
(?<="|')(?:[^"\\]|\\.)*(?="|')
" It\'s big \"problem "
match result:
It\'s big \"problem
("|')(?:[^"\\]|\\.)*("|')
" It\'s big \"problem "
match result:
" It\'s big \"problem "
Messed around at regexpal and ended up with this regex: (Don't ask me how it works, I barely understand even tho I wrote it lol)
"(([^"\\]?(\\\\)?)|(\\")+)+"