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
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.
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
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
I am trying to make a regex which will match any string which looks like this:
User<spaces><Any positive integer here><spaces>Status:<anything here>
Sample expression - User 1 Status: Not Ready.
Regex pattern - ^[User].*\d+.*[Status:].*$
As you can see, I am using ".*" to incorrectly match spaces. I tried to use \s and [" "] instead, but they did not work. How do I handle spaces or tabs in this regex ?
By the way, I am using https://regex101.com/ with JavaScript regex parser to validate my Regex. I don't know if there is any nice regex helper website just for Java and not JavaScript.
Thanks.
You are using character classes (those things surrounded by []) inappropriately. The []s don't mean "match these characters literally". They mean "match any one character in this list". For most characters, they themselves mean "match this literally".
Also, you seem to want to match User: in your regex, yet in the example you provided, there is no :, just User. Please decide whether or not you want the :.
\s is indeed used to match whitespace. You thought it didn't work probably because your regex has other mistakes, making the whole thing not match.
A corrected version of your regex:
^User\s*\d+\s*Status:.*$
Demo
I have the following regex:
^(?=\w+)(-\w+)(?!\.)
Which I'm attempting to match against the following text:
www-test1.examples.com
The regex should match only the -test1 part of the string and only if it is before the first .and after the start of the expression. www can be any string but it should not be matched.
My pattern is not matching the -test1 part. What am I missing?
Java is one of the only languages that support non-fixed-length look-behinds (which basically means you can use quantifiers), so you can technically use the following:
(?<=^\w+)(-\w+)
This will match for -test without capturing the preceding stuff. However, it's generally not advisable to use non-fixed-length look-behinds, as they are not perfect, nor are they very efficient, nor are they portable across other languages. Having said that.. this is a simple pattern, so if you don't care about portability, sure, go for it.
The better solution though is to group what you want to capture, and reference the captured group (in this case, group 1):
^\w+(-\w+)
p.s. - \w will not match a dot, so no need to look ahead for it.
p.p.s. - to answer your question about why your original pattern ^(?=\w+)(-\w+)(?!\.) doesn't match. There are 2 reasons:
1) you start out with a start of string assertion, and then use a lookahead to see if what follows is one or more word chars. But lookaheads are zero-width assertions, meaning no characters are actually consumed in the match, so the pointer doesn't move forward to the next chars after the match. So it sees that "www" matches it, and moves on to the next part of the pattern, but the actual pointer hasn't moved past the start of string. So, it next tries to match your (-\w+) part. Well your string doesn't start with "-" so the pattern fails.
2) (?!\.) is a negative lookahead. Well your example string shows a dot as the very next thing after your "-test" part. So even if #1 didn't fail it, this would fail it.
The problem you're having is the lookahead. In this case, it's inappropriate if you want to capture what's between the - and the first .. The pattern you want is something like this:
(-\w+)(?=\.)
In this case, the contents of capture group 1 will contain the text you want.
Demo on Regex101
Try this:
(?<=www)\-\w+(?=\.)
Demo: https://regex101.com/r/xEpno7/1
This is maybe the 100+1 question regarding regex optional suffixes on SO, but I didn't find any, that could help me :(
I need to extract a part of string from the common pattern:
prefix/s/o/m/e/t/h/i/n/g/suffix
using a regular expression. The prefix is constant and the suffix may not appear at all, so prefix/(.+)/suffix doesn't meet my requirements. Pattern prefix/(.+)(?:/suffix)? returns s/o/m/e/t/h/i/n/g/suffix. The part (?:/suffix)? must be somehow more greedy.
I want to get s/o/m/e/t/h/i/n/g from these input strings:
prefix/s/o/m/e/t/h/i/n/g/suffix
prefix/s/o/m/e/t/h/i/n/g/
prefix/s/o/m/e/t/h/i/n/g
Thanks in advance!
Try
prefix\/(.+?)\/?(?:suffix|$)
The regex need to know when the match is done, so match either suffix or end of line ($), and make the capture non greedy.
See it here at regex101.
Try prefix(.*?)(?:/?(?:suffix|$)) if there are characters allowed before prefix of after suffix.
This requires the match to be as short as possible (reluctant quantifier) and be preceeded by one of 3 things: a single slash right before the end of the input, /suffix or the end of the input. That would match /s/o/m/e/t/h/i/n/g in the test cases you provided but would match more for input like prefix/s/o/m/e/t/h/i/n/g/suff (which is ok IMO since you don't know whether /suff is meant to be part of the match or a typo in the suffix).
I have been trying to resolve this for the past 2 days...
Please help me in understanding why this is happening. My intention is to just select the <HDR> that has a <DTL1 val="92">.....</HDR>
This is my regular expression
(?<=<HDR>).*?<DTL1\sval="3".*?</HDR>
And the input string is:
<HDR>abc<DTL1 val="1"><DTL2 val="2"></HDR><HDR><DTL1 val="92"><DTL2 val="55"></HDR><HDR><DTL1 val="3"><DTL2 val="4"></HDR>
But this regular expression selects
abc<DTL1 val="1"><DTL2 val="2"></HDR><HDR><DTL1 val="92"><DTL2 val="55"></HDR>
Can anyone please help me?
A regex engine will give you always the leftmost match in a string (even if you use a non-greedy quantifier). This is exactly what you obtain.
So, a solution is to forbid the presence of another <HDR> in the parts described by .*? that is too permissive.
You have two technics to do that, you can replace the .*? with:
(?>[^<]+|<(?!/HDR))*
or with:
(?:(?!</HDR).)*+
Most of the time, the first technic is more performant, but if your string contains an high density of <, the second way can give good results too.
The use of a possessive quantifier or an atomic group can reduce the number of steps to obtain a result in particular when the subpattern fails.
Example:
With the first way:
(?<=<HDR>)(?>[^<]+|<(?!/HDR))*<DTL1\sval="3"(?>[^<]+|<(?!/HDR))*</HDR>
or this variant:
(?<=<HDR>)(?:[^<]+|<(?!/HDR|DTL1))*+<DTL1\sval="3"(?:[^<]+|<(?!/HDR))*+</HDR>
With the second way:
(?<=<HDR>)(?:(?!</HDR).)*<DTL1\sval="3"(?:(?!</HDR).)*+</HDR>
or this variant:
(?<=<HDR>)(?:(?!</HDR|DTL1).)*+<DTL1\sval="3"(?:(?!</HDR).)*+</HDR>
Casimir et Hippolyte already gave you a couple of good solutions. I want to elaborate on a few things.
First, why your regex fails to do what you want: (?<=<HDR>).*? tells it to match any number of characters starting with the first character preceded by <HDR>, until it encounters what follows the non-greedy quantifier (<DTL1...). Well, the first character that's preceded by <HDR> is the first a, so it matches everything starting from there until the fixed string <DTL1\sval="3" is encountered.
Casimir et Hippolyte's solutions are for the generalized case, where the contents of the <HDR> tags can be anything other than nested <HDR>'s. You could also do it with a positive look-ahead:
(?<=<HDR>)(.(?!</HDR>))*<DTL1\sval="3".*?</HDR>
However, if the string is guaranteed to be in the structure shown, where the <HDR> tags only contain one or more <DTL1 val="##"> tags, so you know there won't be any closing tags within, you could do it more efficiently by replacing the first .*? with [^/]*:
(?<=<HDR>)[^/]*<DTL1\sval="3".*?</HDR>
A negated character class is more efficient than a zero-width assertion, and if you're using a negated character class, a greedy quantifier becomes more efficient than a lazy one.
Note also that by using a lookbehind to match the opening <HDR>, you're excluding it from the match, but you're including the closing </HDR>. Are you sure that's what you want? You're matching this...
<DTL1 val="3"><DTL2 val="4"></HDR>
...when presumably you want this...
<HDR><DTL1 val="3"><DTL2 val="4"></HDR>
...or this...
<DTL1 val="3"><DTL2 val="4">
So, in the fist case, don't use a lookbehind for the opening tag:
<HDR>(.(?!</HDR>))*<DTL1\sval="3".*?</HDR>
<HDR>[^/]*<DTL1\sval="3".*?</HDR>
In the second case, use a look-ahead for the closing tag:
(?<=<HDR>)(.(?!</HDR>))*<DTL1\sval="3".*?(?=</HDR>)
(?<=<HDR>)[^/]*<DTL1\sval="3".*?(?=</HDR>)