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 do the follow:
I have some strings that I need to separate, they have this form:
node:info:sequence(id:ASDF,LMD)
node:info:sequence:id:QWES
Those are the possible individual string formats...
Now I have to separate them when come concatenated by comma... like this
node:info:sequence(id:ASDF,LMD),node:info:sequence:id:QWES
So I tried
entries.split(",node");
Which... kinda works but of course I cut the "node" part from the previous string, is there anyway I can detect that , followed by node but split it by the comma , only?
You may use
s.split(",(?=node\\b)")
See the regex demo
The positive lookahead (?=node\b) will make sure only those commas are matched that are followed with a whole word node (as the \b is a word boundary).
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
I have some larger text which in essence looks like this:
abc12..manycharshere...hi - abc23...manyothercharshere...jk
Obviously there are two items, each starting with "abc", the numbers (12 and 23) are interesting as well as the "hi" and "jk" at the end.
I would like to create a regular expression which allows me to parse out the numbers, but only if the two characters at the end match, i.e. I am looking for the number related to "jk", but the following regular expression matches the whole string and thus returns "12", not "23" even when non-greedy matching the area with the following:
abc([0-9]+).*?jk
Is there a way to construct a regular expression which matches text like the one above, i.e. retrieving "23" for items ending in "jk"?
Basically I would need something like "match abc followed by a number, but only if there is "jk" at the end before another instance of "abc followed by a number appears"
Note: the texts/matches are an abstraction here, the actual text is more complicated, espially the things that can appear as "manyothercharactershere", I simplified to show the underlying problem more clearly.
Use a regex like this. .*abc([0-9]+).*?jk
demo here
I think you want something like this,
abc([0-9]+)(?=(?:(?!jk|abc[0-9]).)*jk)
DEMO
You need to use negative lookahead here to make it work:
abc(?!.*?abc)([0-9]+).*?jk
RegEx Demo
Here (?!.*?abc) is negative lookahead that makes sure to match abc where it is NOT followed by another abc thus making sure closes string between abc and jk is matched.
Being non-greedy does not change the rule, that the first match is returned. So abc([0-9]+).*?jk will find the first jk after “abcnumber” rather than the last one, but still match the first “abcnumber”.
One way to solve this is to tell that the dot should not match abc([0-9]+):
abc([0-9]+)((?!abc([0-9]+)).)*jk
If it is not important to have the entire pattern being an exact match you can do it simpler:
.*(abc([0-9]+).*?jk)
In this case, it’s group 1 which contains your intended match. The pattern uses a greedy matchall to ensure that the last possible “abcnumber” is matched within the group.
Assuming that hyphen separates "items", this regex will capture the numbers from the target item:
abc([0-9]+)[^-]*?jk
See demo
I'm not sure if this is a regex question, but i need to be able to grab whats inside single qoutes, and surround them with something. For example:
this is a 'test' of 'something' and 'I' 'really' want 'to' 'solve this'
would turn into
this is a ('test') of ('something') and ('I') ('really') want ('to') ('solve this')
Any help you could provide would be great!
Thanks!
String str = "this is a 'test' of 'something'";
String rep = str.replaceAll("'[^']*'", "($0)"); // stand back, I know regex
What I did here is use toe replaceAll() method which searches for all matches for regex "'[^']*'" and replaces them with regex "($0)".
The pattern "'[^']*'" matches all substrings that start and end with a single quote ('), and between them are any characters, except another single quote ([^']), and those can appear any number of times (*). Replacing those with "($0)" means taking every match ($0) and wrapping it with parenthesis.
One easy way (but not always valid) is the following
If always you have [ '] and [' ] , you can do this:
myString.replace(" '"," ('"); // replaces all <space_apostrophe> with <space_bracket_apostrophe>
Do the same for the rear bracket :)
One more thing - why do you even have apostrophes-surrounded words? Is it a must that they must be like that? If you made them like that, why did you do it and then look for another approach !
If you can ignore single apostrophes, you could do like this (C# code, should be easy to translate)
string input = "this is a 'test' of 'something' and ...";
Console.WriteLine(Regex.Replace(input, "'([^']*)'", "('$1')"));