I am having a string say
my $str = "FILLER-1-1,EQPT:MN,EQPT_MISSING,NSA,04-30,15-07-13,NEND,NA";
I want to match a pattern say
my $pattern = "FILLER-1-1";
I am using the below regexp
$reg = $str =~ /$pattern/;
This is working fine
Now the problem is it is also matching if our string is
FILLER-1-10/FILLER-1-11/FILLER-1-12 so on ...
I dont want to match this. Also I don't want my regexp to be like
$reg = $str =~ /$pattern\W+/;
This one is working against the above mentioned issue but \W may come or not come. In some strings it can come while in other it may not come. So i need the regexp to match only FILLER-1-1 without using \W+ and it should match specifically FILLER-1-10
Note: If somebody is doing -(minus) rating to my question, please let me know what's wrong in the code. It will be appreciable if the person write the comment too
As \w matches [a-zA-Z0-9], you can use the zero-width assumption \b, which denotes a change in \w state (called a "word boundary", hence the "b" shortcut):
/FILLER-1-1\b/
This means that there needs to be a character that differs from the previous word state - a word state change.
It will match
FILLER-1-1.
FILLER-1-1&
FILLER-1-1,
It will not match
FILLER-1-1a
FILLER-1-16
Read more about it here.
If you want to match FILLER at the start of the input (line) followed by two numbers, this simple regex should work:
/~FILLER-\d+-\d+/
~ matches the beginning of the input
\d matches any digit ([0-9])
+ matches at least one, but can match any number
use ? quantifier like so:
/FILLER-\d-\d\W?/
The \W? means not a word zero or one time
I've got a regular expression with capture groups that matches what I want in a broader context. I then take capture group $1 and use it for my needs. That's easy.
But how to use capture groups with s/// when I just want to replace the content of $1, not the entire regex, with my replacement?
For instance, if I do:
$str =~ s/prefix (something) suffix/42/
prefix and suffix are removed. Instead, I would like something to be replaced by 42, while keeping prefix and suffix intact.
As I understand, you can use look-ahead or look-behind that don't consume characters. Or save data in groups and only remove what you are looking for. Examples:
With look-ahead:
s/your_text(?=ahead_text)//;
Grouping data:
s/(your_text)(ahead_text)/$2/;
If you only need to replace one capture then using #LAST_MATCH_START and #LAST_MATCH_END (with use English; see perldoc perlvar) together with substr might be a viable choice:
use English qw(-no_match_vars);
$your_string =~ m/aaa (bbb) ccc/;
substr $your_string, $LAST_MATCH_START[1], $LAST_MATCH_END[1] - $LAST_MATCH_START[1], "new content";
# replaces "bbb" with "new content"
This is an old question but I found the below easier for replacing lines that start with >something to >something_else. Good for changing the headers for fasta sequences
while ($filelines=~ />(.*)\s/g){
unless ($1 =~ /else/i){
$filelines =~ s/($1)/$1\_else/;
}
}
I use something like this:
s/(?<=prefix)(group)(?=suffix)/$1 =~ s|text|rep|gr/e;
Example:
In the following text I want to normalize the whitespace but only after ::=:
some text := a b c d e ;
Which can be achieved with:
s/(?<=::=)(.*)/$1 =~ s|\s+| |gr/e
Results with:
some text := a b c d e ;
Explanation:
(?<=::=): Look-behind assertion to match ::=
(.*): Everything after ::=
$1 =~ s|\s+| |gr: With the captured group normalize whitespace. Note the r modifier which makes sure not to attempt to modify $1 which is read-only. Use a different sub delimiter (|) to not terminate the replacement expression.
/e: Treat the replacement text as a perl expression.
Use lookaround assertions. Quoting the documentation:
Lookaround assertions are zero-width patterns which match a specific pattern without including it in $&. Positive assertions match when their subpattern matches, negative assertions match when their subpattern fails. Lookbehind matches text up to the current match position, lookahead matches text following the current match position.
If the beginning of the string has a fixed length, you can thus do:
s/(?<=prefix)(your capture)(?=suffix)/$1/
However, ?<= does not work for variable length patterns (starting from Perl 5.30, it accepts variable length patterns whose length is smaller than 255 characters, which enables the use of |, but still prevents the use of *). The work-around is to use \K instead of (?<=):
s/.*prefix\K(your capture)(?=suffix)/$1/
I have this input:
AB2.HYNN.KABCDSEG.L000.G0001V00
AB2.HYNN.GABCDSEG.L000.G0005V00
I would like to remove all which finish by GXXXXVXX in the string.
When i use this code:
$result =~ s/\.G.*V.*$//g;
print "$result \n";
The result is :
AB2.HYNN.KABCDSEG.L000
AB2.HYNN
It seems each time the regex find ".G" it removes with blank .
I don't understand.
I would like to have this:
AB2.HYNN.KABCDSEG.L000
AB2.HYNN.GABCDSEG.L000
How i can do this in regex ?
Update:
After talking in the comments, the final solution was:
s/\.G\w+V\w+$//;
In your regex:
s/\.G.*V.*$//g;
those .* are greedy and will match as much as possible. The only requirement you have is that there must be a V after the .G somewhere, so it will truncate the string from the first .G it finds, as long as it is followed by a V. There is no need for the /g modifier here, because any match that occurs will delete the rest of the string. Unless you have newlines, because . does not match newlines without the /s modifier.
$result =~ s/\.G\d+V\d+//g;
Works on given input.
I would like some help in getting this regex to accept the space character.
The following regex works ^a|a$|a but this one doesn't ^tips to|tips to$|tips to.
Space is just as-is in a regex (you just put the space character, that should work). Alternatively you can use \s special character. For example, in Perl:
my $test = "Helloworld";
if ($test =~ m/ /)
{
print("Has space\n");
}
Also if you can specify more what you want to use the regex for, we might be able to help better.
try escaping just the last space (the regex engine will then "see" that "tips to" is one block - at least for the last OR)
^tips to|tips to$|tips\ to
or to be on the safe side group what your searching for
^(tips to)|(tips to)$|(tips to)
[EDIT 1]
so here's the solution the OP is using:
^"tips to"|"tips to"$|"tips to"
The regular expression that matches 1 space character is 1 space character.
I have a value like this:
"Foo Bar" "Another Value" something else
What regex will return the values enclosed in the quotation marks (e.g. Foo Bar and Another Value)?
In general, the following regular expression fragment is what you are looking for:
"(.*?)"
This uses the non-greedy *? operator to capture everything up to but not including the next double quote. Then, you use a language-specific mechanism to extract the matched text.
In Python, you could do:
>>> import re
>>> string = '"Foo Bar" "Another Value"'
>>> print re.findall(r'"(.*?)"', string)
['Foo Bar', 'Another Value']
I've been using the following with great success:
(["'])(?:(?=(\\?))\2.)*?\1
It supports nested quotes as well.
For those who want a deeper explanation of how this works, here's an explanation from user ephemient:
([""']) match a quote; ((?=(\\?))\2.) if backslash exists, gobble it, and whether or not that happens, match a character; *? match many times (non-greedily, as to not eat the closing quote); \1 match the same quote that was use for opening.
I would go for:
"([^"]*)"
The [^"] is regex for any character except '"'
The reason I use this over the non greedy many operator is that I have to keep looking that up just to make sure I get it correct.
Lets see two efficient ways that deal with escaped quotes. These patterns are not designed to be concise nor aesthetic, but to be efficient.
These ways use the first character discrimination to quickly find quotes in the string without the cost of an alternation. (The idea is to discard quickly characters that are not quotes without to test the two branches of the alternation.)
Content between quotes is described with an unrolled loop (instead of a repeated alternation) to be more efficient too: [^"\\]*(?:\\.[^"\\]*)*
Obviously to deal with strings that haven't balanced quotes, you can use possessive quantifiers instead: [^"\\]*+(?:\\.[^"\\]*)*+ or a workaround to emulate them, to prevent too much backtracking. You can choose too that a quoted part can be an opening quote until the next (non-escaped) quote or the end of the string. In this case there is no need to use possessive quantifiers, you only need to make the last quote optional.
Notice: sometimes quotes are not escaped with a backslash but by repeating the quote. In this case the content subpattern looks like this: [^"]*(?:""[^"]*)*
The patterns avoid the use of a capture group and a backreference (I mean something like (["']).....\1) and use a simple alternation but with ["'] at the beginning, in factor.
Perl like:
["'](?:(?<=")[^"\\]*(?s:\\.[^"\\]*)*"|(?<=')[^'\\]*(?s:\\.[^'\\]*)*')
(note that (?s:...) is a syntactic sugar to switch on the dotall/singleline mode inside the non-capturing group. If this syntax is not supported you can easily switch this mode on for all the pattern or replace the dot with [\s\S])
(The way this pattern is written is totally "hand-driven" and doesn't take account of eventual engine internal optimizations)
ECMA script:
(?=["'])(?:"[^"\\]*(?:\\[\s\S][^"\\]*)*"|'[^'\\]*(?:\\[\s\S][^'\\]*)*')
POSIX extended:
"[^"\\]*(\\(.|\n)[^"\\]*)*"|'[^'\\]*(\\(.|\n)[^'\\]*)*'
or simply:
"([^"\\]|\\.|\\\n)*"|'([^'\\]|\\.|\\\n)*'
Peculiarly, none of these answers produce a regex where the returned match is the text inside the quotes, which is what is asked for. MA-Madden tries but only gets the inside match as a captured group rather than the whole match. One way to actually do it would be :
(?<=(["']\b))(?:(?=(\\?))\2.)*?(?=\1)
Examples for this can be seen in this demo https://regex101.com/r/Hbj8aP/1
The key here is the the positive lookbehind at the start (the ?<= ) and the positive lookahead at the end (the ?=). The lookbehind is looking behind the current character to check for a quote, if found then start from there and then the lookahead is checking the character ahead for a quote and if found stop on that character. The lookbehind group (the ["']) is wrapped in brackets to create a group for whichever quote was found at the start, this is then used at the end lookahead (?=\1) to make sure it only stops when it finds the corresponding quote.
The only other complication is that because the lookahead doesn't actually consume the end quote, it will be found again by the starting lookbehind which causes text between ending and starting quotes on the same line to be matched. Putting a word boundary on the opening quote (["']\b) helps with this, though ideally I'd like to move past the lookahead but I don't think that is possible. The bit allowing escaped characters in the middle I've taken directly from Adam's answer.
The RegEx of accepted answer returns the values including their sourrounding quotation marks: "Foo Bar" and "Another Value" as matches.
Here are RegEx which return only the values between quotation marks (as the questioner was asking for):
Double quotes only (use value of capture group #1):
"(.*?[^\\])"
Single quotes only (use value of capture group #1):
'(.*?[^\\])'
Both (use value of capture group #2):
(["'])(.*?[^\\])\1
-
All support escaped and nested quotes.
I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:
(['"])(?:(?!\1|\\).|\\.)*\1
It does the trick and is still pretty simple and easy to maintain.
Demo (with some more test-cases; feel free to use it and expand on it).
PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:
(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)
Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.
Alternatively, modify the initial version by simply adding a group and extract the string form $2:
(['"])((?:(?!\1|\\).|\\.)*)\1
PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.
A very late answer, but like to answer
(\"[\w\s]+\")
http://regex101.com/r/cB0kB8/1
The pattern (["'])(?:(?=(\\?))\2.)*?\1 above does the job but I am concerned of its performances (it's not bad but could be better). Mine below it's ~20% faster.
The pattern "(.*?)" is just incomplete. My advice for everyone reading this is just DON'T USE IT!!!
For instance it cannot capture many strings (if needed I can provide an exhaustive test-case) like the one below:
$string = 'How are you? I\'m fine, thank you';
The rest of them are just as "good" as the one above.
If you really care both about performance and precision then start with the one below:
/(['"])((\\\1|.)*?)\1/gm
In my tests it covered every string I met but if you find something that doesn't work I would gladly update it for you.
Check my pattern in an online regex tester.
This version
accounts for escaped quotes
controls backtracking
/(["'])((?:(?!\1)[^\\]|(?:\\\\)*\\[^\\])*)\1/
MORE ANSWERS! Here is the solution i used
\"([^\"]*?icon[^\"]*?)\"
TLDR;
replace the word icon with what your looking for in said quotes and voila!
The way this works is it looks for the keyword and doesn't care what else in between the quotes.
EG:
id="fb-icon"
id="icon-close"
id="large-icon-close"
the regex looks for a quote mark "
then it looks for any possible group of letters thats not "
until it finds icon
and any possible group of letters that is not "
it then looks for a closing "
I liked Axeman's more expansive version, but had some trouble with it (it didn't match for example
foo "string \\ string" bar
or
foo "string1" bar "string2"
correctly, so I tried to fix it:
# opening quote
(["'])
(
# repeat (non-greedy, so we don't span multiple strings)
(?:
# anything, except not the opening quote, and not
# a backslash, which are handled separately.
(?!\1)[^\\]
|
# consume any double backslash (unnecessary?)
(?:\\\\)*
|
# Allow backslash to escape characters
\\.
)*?
)
# same character as opening quote
\1
string = "\" foo bar\" \"loloo\""
print re.findall(r'"(.*?)"',string)
just try this out , works like a charm !!!
\ indicates skip character
My solution to this is below
(["']).*\1(?![^\s])
Demo link : https://regex101.com/r/jlhQhV/1
Explanation:
(["'])-> Matches to either ' or " and store it in the backreference \1 once the match found
.* -> Greedy approach to continue matching everything zero or more times until it encounters ' or " at end of the string. After encountering such state, regex engine backtrack to previous matching character and here regex is over and will move to next regex.
\1 -> Matches to the character or string that have been matched earlier with the first capture group.
(?![^\s]) -> Negative lookahead to ensure there should not any non space character after the previous match
Unlike Adam's answer, I have a simple but worked one:
(["'])(?:\\\1|.)*?\1
And just add parenthesis if you want to get content in quotes like this:
(["'])((?:\\\1|.)*?)\1
Then $1 matches quote char and $2 matches content string.
All the answer above are good.... except they DOES NOT support all the unicode characters! at ECMA Script (Javascript)
If you are a Node users, you might want the the modified version of accepted answer that support all unicode characters :
/(?<=((?<=[\s,.:;"']|^)["']))(?:(?=(\\?))\2.)*?(?=\1)/gmu
Try here.
echo 'junk "Foo Bar" not empty one "" this "but this" and this neither' | sed 's/[^\"]*\"\([^\"]*\)\"[^\"]*/>\1</g'
This will result in: >Foo Bar<><>but this<
Here I showed the result string between ><'s for clarity, also using the non-greedy version with this sed command we first throw out the junk before and after that ""'s and then replace this with the part between the ""'s and surround this by ><'s.
From Greg H. I was able to create this regex to suit my needs.
I needed to match a specific value that was qualified by being inside quotes. It must be a full match, no partial matching could should trigger a hit
e.g. "test" could not match for "test2".
reg = r"""(['"])(%s)\1"""
if re.search(reg%(needle), haystack, re.IGNORECASE):
print "winning..."
Hunter
If you're trying to find strings that only have a certain suffix, such as dot syntax, you can try this:
\"([^\"]*?[^\"]*?)\".localized
Where .localized is the suffix.
Example:
print("this is something I need to return".localized + "so is this".localized + "but this is not")
It will capture "this is something I need to return".localized and "so is this".localized but not "but this is not".
A supplementary answer for the subset of Microsoft VBA coders only one uses the library Microsoft VBScript Regular Expressions 5.5 and this gives the following code
Sub TestRegularExpression()
Dim oRE As VBScript_RegExp_55.RegExp '* Tools->References: Microsoft VBScript Regular Expressions 5.5
Set oRE = New VBScript_RegExp_55.RegExp
oRE.Pattern = """([^""]*)"""
oRE.Global = True
Dim sTest As String
sTest = """Foo Bar"" ""Another Value"" something else"
Debug.Assert oRE.test(sTest)
Dim oMatchCol As VBScript_RegExp_55.MatchCollection
Set oMatchCol = oRE.Execute(sTest)
Debug.Assert oMatchCol.Count = 2
Dim oMatch As Match
For Each oMatch In oMatchCol
Debug.Print oMatch.SubMatches(0)
Next oMatch
End Sub