I want to split a string like this:
abc//def//ghi
into a part before and after the first occurrence of //:
a: abc
b: //def//ghi
I'm currently using this regex:
(?<a>.*?)(?<b>//.*)
Which works fine so far.
However, sometimes the // is missing in the source string and obviously the regex fails to match. How is it possible to make the second group optional?
An input like abc should be matched to:
a: abc
b: (empty)
I tried (?<a>.*?)(?<b>//.*)? but that left me with lots of NULL results in Expresso so I guess it's the wrong idea.
Try a ^ at the begining of your expression to match the begining of the string and a $ at the end to match the end of the string (this will make the ungreedy match work).
^(?<a>.*?)(?<b>//.*)?$
A proof of Stevo3000's answer (Python):
import re
test_strings = ['abc//def//ghi', 'abc//def', 'abc']
regex = re.compile("(?P<a>.*?)(?P<b>//.*)?$")
for ts in test_strings:
match = regex.match(ts)
print 'a:', match.group('a'), 'b:', match.group('b')
a: abc b: //def//ghi
a: abc b: //def
a: abc b: None
Why use group matching at all? Why not just split by "//", either as a regex or a plain string?
use strict;
my $str = 'abc//def//ghi';
my $short = 'abc';
print "The first:\n";
my #groups = split(/\/\//, $str, 2);
foreach my $val (#groups) {
print "$val\n";
}
print "The second:\n";
#groups = split(/\/\//, $short, 2);
foreach my $val (#groups) {
print "$val\n";
}
gives
The first:
abc
def//ghi
The second:
abc
[EDIT: Fixed to return max 2 groups]
Related
I have a string that looks like this, they are ids in a table:
1,2,3,4,5,6,7,8,9
If someone deletes something from the database, I will need to update the string. I know that doing this it will remove the value, but not the commas. Any idea how can I check if the id has a comma before and after so my string doesn't break?
$new_values = $original_values[0];
$new_values =~ s/$car_id//;
Result: 1,2,,4,5,6,7,8,9 using the above sample (bad). It should be 1,2,4,5,6,7,8,9.
To remove the $car_id from the string:
my $car_id = 3;
my $new_values = q{1,2,3,4,5,6,7,8,9};
$new_values = join q{,}, grep { $_ != $car_id }
split /,/, $new_values;
say $new_values;
# Prints:
# 1,2,4,5,6,7,8,9
If you already removed the id(s), and you need to remove the extra commas, reformat the string like so:
my $new_values = q{,,1,2,,4,5,6,7,8,9,,,};
$new_values = join q{,}, grep { /\d/ } split /,/, $new_values;
say $new_values;
# Prints:
# 1,2,4,5,6,7,8,9
You can use
s/^$car_id,|,$car_id\b//
Details
^ - start of string
$car_id - variable value
, - comma
| - or
, - comma
$car_id - variable value
\b - word boundary.
s/^\Q$car_id\E,|,\Q$car_id\E\b//
Another approach is to store an extra leading and trailing comma (,1,2,3,4,5,6,7,8,9,)
The main benefit is that it makes it easier to search for the id using SQL (since you can search for ,$car_id,). Same goes for editing it.
On the Perl side, you'd use
s/,\K\Q$car_id\E,// # To remove
substr($_, 1, -1) # To get actual string
Ugly way: use regex to remove the value, then simplify
$new_values = $oringa_value[0];
$new_values =~ s/$car_id//;
$new_values =~ s/,+/,/;
Nice way: split and merge
$new_values = $oringa_value[0];
my #values = split(/,/, $new_values);
my $index = 0;
$index++ until $values[$index] eq $car_id;
splice(#values, $index, 1);
$new_values = join(',', #values);
I recently came across a puzzle to find a regular expression that matches:
5-character-long strings comprised of lowercase English letters in ascending ASCII order
Valid examples include:
aaaaa
abcde
xxyyz
ghost
chips
demos
Invalid examples include:
abCde
xxyyzz
hgost
chps
My current solution is kludgy. I use the regex:
(?=^[a-z]{5}$)^(a*b*c*d*e*f*g*h*i*j*k*l*m*n*o*p*q*r*s*t*u*v*w*x*y*z*)$
which uses a non-consuming capture group to assert a string length of 5, and then verifies that the string comprises of lowercase English letters in order (see Rubular).
Instead, I'd like to use back references inside character classes. Something like:
^([a-z])([\1-z])([\2-z])([\3-z])([\4-z])$
The logic for the solution (see Rubular) in my head is to capture the first character [a-z], use it as a backrefence in the second character class and so on. However, \1, \2 ... within character classes seem to refer to ASCII values of 1, 2... effectively matching any four- or five-character string.
I have 2 questions:
Can I use back references in my character classes to check for ascending order strings?
Is there any less-hacky solution to this puzzle?
I'm posting this answer more as a comment than an answer since it has better formatting than comments.
Related to your questions:
Can I use back references in my character classes to check for ascending order strings?
No, you can't. If you take a look a backref regular-expressions section, you will find below documentation:
Parentheses and Backreferences Cannot Be Used Inside Character Classes
Parentheses cannot be used inside character classes, at least not as metacharacters. When you put a parenthesis in a character class, it is treated as a literal character. So the regex [(a)b] matches a, b, (, and ).
Backreferences, too, cannot be used inside a character class. The \1 in a regex like (a)[\1b] is either an error or a needlessly escaped literal 1. In JavaScript it's an octal escape.
Regarding your 2nd question:
Is there any less-hacky solution to this puzzle?
Imho, your regex is perfectly well, you could shorten it very little at the beginning like this:
(?=^.{5}$)^a*b*c*d*e*f*g*h*i*j*k*l*m*n*o*p*q*r*s*t*u*v*w*x*y*z*$
^--- Here
Regex demo
If you are willing to use Perl (!), this will work:
/^([a-z])((??{"[$1-z]"}))((??{"[$2-z]"}))((??{"[$3-z]"}))(??{"[$4-z]"})$/
Since someone has broken the ice by using Perl, this is a
Perl solution I guess ..
Note that this is a basic non-regex solution that just happens to be
stuffed into code constructs inside a Perl regex.
The interesting thing is that if a day comes when you need the synergy
of regex/code this is a good choice.
It is possible then that instead of a simple [a-z] character, you may
use a very complex pattern in it's place and using a check vs. last.
That is power !!
The regex ^(?:([a-z])(?(?{ $last gt $1 })(?!)|(?{ $last = $1 }))){5}$
Perl code
use strict;
use warnings;
$/ = "";
my #DAry = split /\s+/, <DATA>;
my $last;
for (#DAry)
{
$last = '';
if (
/
^ # BOS
(?: # Cluster begin
( [a-z] ) # (1), Single a-z letter
# Code conditional
(?(?{
$last gt $1 # last > current ?
})
(?!) # Fail
| # else,
(?{ $last = $1 }) # Assign last = current
)
){5} # Cluster end, do 5 times
$ # EOS
/x )
{
print "good $_\n";
}
else {
print "bad $_\n";
}
}
__DATA__
aaaaa
abcde
xxyyz
ghost
chips
demos
abCde
xxyyzz
hgost
chps
Output
good aaaaa
good abcde
good xxyyz
good ghost
good chips
good demos
bad abCde
bad xxyyzz
bad hgost
bad chps
Ah, well, it's a finite set, so you can always enumerate it with alternation! This emits a "brute force" kind of regex in a little perl REPL:
#include <stdio.h>
int main(void) {
printf("while (<>) { if (/^(?:");
for (int a = 'a'; a <= 'z'; ++a)
for (int b = a; b <= 'z'; ++b)
for (int c = b; c <= 'z'; ++c) {
for (int d = c; d <= 'y'; ++d)
printf("%c%c%c%c[%c-z]|", a, b, c, d, d);
printf("%c%c%czz", a, b, c);
if (a != 'z' || b != 'z' || c != 'z') printf("|\n");
}
printf(")$/x) { print \"Match!\\n\" } else { print \"No match.\\n\" }}\n");
return 0;
}
And now:
$ gcc r.c
$ ./a.out > foo.pl
$ cat > data.txt
aaaaa
abcde
xxyyz
ghost
chips
demos
abCde
xxyyzz
hgost
chps
^D
$ perl foo.pl < data.txt
Match!
Match!
Match!
Match!
Match!
Match!
No match.
No match.
No match.
No match.
The regex is only 220Kb or so ;-)
I have a string that looks like this:
my_str = "This sentence has a [b|bolded] word, and [b|another] one too!"
And I need it to be converted into this:
new_str = "This sentence has a <b>bolded</b> word, and <b>another</b> one too!"
Is it possible to use Python's string.replace or re.sub method to do this intelligently?
Just capture all the characters before | inside [] into a group . And the part after | into another group. Just call the captured groups through back-referencing in the replacement part to get the desired output.
Regex:
\[([^\[\]|]*)\|([^\[\]]*)\]
Replacemnet string:
<\1>\2</\1>
DEMO
>>> import re
>>> s = "This sentence has a [b|bolded] word, and [b|another] one too!"
>>> m = re.sub(r'\[([^\[\]|]*)\|([^\[\]]*)\]', r'<\1>\2</\1>', s)
>>> m
'This sentence has a <b>bolded</b> word, and <b>another</b> one too!'
Explanation...
Try this expression: [[]b[|](\w+)[]] shorter version can also be \[b\|(\w+)\]
Where the expression is searching for anything that starts with [b| captures what is between it and the closing ] using \w+ which means [a-zA-Z0-9_] to include a wider range of characters you can also use .*? instead of \w+ which will turn out in \[b\|(.*?)\]
Online Demo
Sample Demo:
import re
p = re.compile(ur'[[]b[|](\w+)[]]')
test_str = u"This sentence has a [b|bolded] word, and [b|another] one too!"
subst = u"<bold>$1</bold>"
result = re.sub(p, subst, test_str)
Output:
This sentence has a <bold>bolded</bold> word, and <bold>another</bold> one too!
Just for reference, in case you don't want two problems:
Quick answer to your particular problem:
my_str = "This sentence has a [b|bolded] word, and [b|another] one too!"
print my_str.replace("[b|", "<b>").replace("]", "</b>")
# output:
# This sentence has a <b>bolded</b> word, and <b>another</b> one too!
This has the flaw that it will replace all ] to </b> regardless whether it is appropriate or not. So you might want to consider the following:
Generalize and wrap it in a function
def replace_stuff(s, char):
begin = s.find("[{}|".format(char))
while begin != -1:
end = s.find("]", begin)
s = s[:begin] + s[begin:end+1].replace("[{}|".format(char),
"<{}>".format(char)).replace("]", "</{}>".format(char)) + s[end+1:]
begin = s.find("[{}|".format(char))
return s
For example
s = "Don't forget to [b|initialize] [code|void toUpper(char const *s)]."
print replace_stuff(s, "code")
# output:
# "Don't forget to [b|initialize] <code>void toUpper(char const *s)</code>."
I want to get an array of all the words with capital letters that are included in the string. But only if the line begins with "set".
For example:
- string "setUserId", result array("User", "Id")
- string "getUserId", result false
Without limitation about "set" RegEx look like /([A-Z][a-z]+)/
$str ='setUserId';
$rep_str = preg_replace('/^set/','',$str);
if($str != $rep_str) {
$array = preg_split('/(?<=[a-z])(?=[A-Z])/',$rep_str);
var_dump($array);
}
See it
Also your regex will also work.:
$str = 'setUserId';
if(preg_match('/^set/',$str) && preg_match_all('/([A-Z][a-z]*)/',$str,$match)) {
var_dump($match[1]);
}
See it
I have something like this
AD ABCDEFG HIJKLMN
AB HIJKLMN
AC DJKEJKW SJKLAJL JSHELSJ
Rule: Always 2 Chars Code (AB|AC|AD) at line beginning then any number of 7 Chars codes following.
With this regex:
^(AB|AC|AD)|((\S{7})?
in this groovy code sample:
def m= Pattern.compile(/^(AB|AC|AD)|((\S{7})?)/).matcher("AC DJKEJKW SJKLAJL JSHELSJ")
println m.getCount()
I always get 8 as count, means it counts the spaces.
How do I get 4 groups (as expected) without spaces ?
Thanks from a not-yet-regex-expert
Sven
Using this code:
def input = [ 'AD ABCDEFG HIJKLMN', 'AB HIJKLMN', 'AC DJKEJKW SJKLAJL JSHELSJ' ]
def regexp = /^(AB|AC|AD)|((\S{7})+)/
def result = input.collect {
matcher = ( it =~ regexp )
println "got $matcher.count for $it"
matcher.collect { it[0] }
}
println result
I get the output
got 3 for AD ABCDEFG HIJKLMN
got 2 for AB HIJKLMN
got 4 for AC DJKEJKW SJKLAJL JSHELSJ
[[AD, ABCDEFG, HIJKLMN], [AB, HIJKLMN], [AC, DJKEJKW, SJKLAJL, JSHELSJ]]
Is this more what you wanted?
This pattern will match your requirements
^A[BCD](?:\s\S{7})+
See it here online on Regexr
Meaning start with A then either a B or a C or a D. This is followed by at least one group consisting of a whitespace followed by 7 non whitespaces.