We Keep Coding

Get what is in string from one quotation mark to other [duplicate]

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

Remove not used Merge tags Regular expression

I am using C# and writing mail merge application. My users store the templates as follows
Dear [[-UserName-]],
You have been subscribed for [[-SubscriptionName-]]...
and so on. There will be lot of custom fields between [[-xxxxx-]] place holders. I am merging them fine. But sometimes they don't pass for some place holders. I would like to find those things using regular expressions and replace them with empty strings.
Technically, I want to find out the regular expression to find [[-what ever it is in between-]] and replace with empty string along with [[--]] tags

You can use:
\[{2}(-.*?-)\]{2}
# look for [[-, -]] and anything in between.
This is called a lazy dot star, see a demo on regex101.com.

Since there are no dashes inside the tag names, you can use this regex:
\[\[-([^-]+)-\]\]
The tag name will be in capture group 1. Obviously, I recommend using the #"" type of string for the regex.
You will want to use the replacement string:
[[--]]
Essentially the regex finds [[-, then captures a bunch of non - characters, then finds -]].

Why does changing this regex class to .+ not provide any match?

If I use this
string showPattern = #"return new_lightox\(this\);"">[a-zA-Z0-9(\s),!\?\-:'&%]+</a>";
MatchCollection showMatches = Regex.Matches(pageSource, showPattern);
I get some matches but I want to get rid of [a-zA-Z0-9(\s),!\?\-:'&%]+and use any char .+
but if do this I get no match at all.
What am I doing wrong?

By default "." does not match newlines, but the class \s does.

To let . match newline, turn on SingleLine/DOTALL mode - either using a flag in the function call (as Abel's answer shows), or using the inline modifier (?s), like this for the whole expression:
"(?s)return new_lightox\(this\);"">.+</a>"
Or for just the specific part of it:
"return new_lightox\(this\);"">(?s:.+)</a>"
It might be better to take that a step further and do this:
"return new_lightox\(this\);"">(?s:(?!</?a).+)</a>"
Which should prevent the closing </a> from belonging to a different link.
However, you need to be very wary here - it's not clear what you're doing overall, but regex is not a good tool for parsing HTML with, and can cause all sorts of problems. Look at using a HTML DOM parser instead, such as HtmlAgilityPack.

You're matching a tag, so you probably want something along these lines, instead of .+:
string showPattern = #"return new_lightox\(this\);"">[^<]+</a>";
The reason that the match doesn't hit is possibly because you are missing the multiline/singleline flag and the closing tag is on the next line. In other words, this should work too:
// SingleLine option changes the dot (.) to match newlines too
MatchCollection showMatches = Regex.Matches(
pageSource,
showPattern,
RegexOptions.SingleLine);

Regular Expression to reject special characters other than commas

I am working in asp.net. I am using Regular Expression Validator
Could you please help me in creating a regular expression for not allowing special characters other than comma. Comma has to be allowed.
I checked in regexlib, however I could not find a match. I treid with ^(a-z|A-Z|0-9)*[^#$%^&*()']*$ . When I add other characters as invalid, it does not work.
Also could you please suggest me a place where I can find a good resource of regular expressions? regexlib seems to be big; but any other place which lists very limited but most used examples?
Also, can I create expressions using C# code? Any articles for that?

[\w\s,]+
works fine, as you can see bellow.
RegExr is a great place to test your regular expressions with real time results, it also comes with a very complete list of common expressions.
[] character class \w Matches any word character (alphanumeric & underscore). \s
Matches any whitespace character (spaces, tabs, line breaks). , include comma + is greedy match; which will match the previous 1 or more times.

[\d\w\s,]*
Just a guess

To answer on any articles, I got started here, find it to be an excellent resource:
http://www.regular-expressions.info/
For your current problem, try something like this:
[\w\s,]*
Here's a breakdown:
Match a single character present in the list below «[\w\s,]*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
A word character (letters, digits, etc.) «\w»
A whitespace character (spaces, tabs, line breaks, etc.) «\s»
The character “,” «,»

For a single character that is not a comma, [^,] should work perfectly fine.

You can try [\w\s,] regular expression. This regex will match only alpha-numeric characters and comma. If any other character appears within text, then this wont match.
For your second question regarding regular expression resource, you can goto
http://www.regular-expressions.info/
This website has lot of tutorials on regex, plus it has lot of usefult information.
Also, can I create expressions using
C# code? Any articles for that?
By this, do you mean to say you want to know which class and methods for regular expression execution? Or you want tool that will create regular expression for you?

You can create expressions with C#, something like this usually does the trick:
Regex regex = new Regex(#"^[a-z | 0-9 | /,]*$", RegexOptions.IgnoreCase);
System.Console.Write("Enter Text");
String s = System.Console.ReadLine();
Match match = regex.Match(s);
if (match.Success == true)
{
System.Console.WriteLine("True");
}
else
{
System.Console.WriteLine("False");
}
System.Console.ReadLine();
You need to import the System.Text.RegularExpressions;
The regular expression above, accepts only numbers, letters (both upper and lower case) and the comma.
For a small introduction to Regular Expressions, I think that the book for MCTS 70-536 can be of a big help, I am pretty sure that you can either download it from somewhere or obtain a copy.
I am assuming that you never messed around with regular expressions in C#, hence I provided the code above.
Hope this helps.

Thank you, all..
[\w\s,]* works
Let me go through regular-expressions.info and come back if I need further support.
Let me try the C# code approach and come back if I need further support.
[This forum is awesome. Quality replies so qucik..]
Thanks again

(…) is denoting a grouping and not a character set that’s denoted with […]. So try this:
^[a-zA-Z0-9,]*$
This will only allow alphanumeric characters and the comma.

How to check if a string starts and ends with specific strings?

I have a string like:
string str = "https://abce/MyTest";
I want to check if the particular string starts with https:// and ends with /MyTest.
How can I acheive that?

This regular expression:
^https://.*/MyTest$
will do what you ask.
^ matches the beginning of the string.
https:// will match exactly that.
.* will match any number of characters (the * part) of any kind (the . part). If you want to make sure there is at least one character in the middle, use .+ instead.
/MyTest matches exactly that.
$ matches the end of the string.
To verify the match, use:
Regex.IsMatch(str, #"^https://.*/MyTest$");
More info at the MSDN Regex page.

Try the following:
var str = "https://abce/MyTest";
var match = Regex.IsMatch(str, "^https://.+/MyTest$");
The ^ identifier matches the start of the string, while the $ identifier matches the end of the string. The .+ bit simply means any sequence of chars (except a null sequence).
You need to import the System.Text.RegularExpressions namespace for this, of course.

I want to check if the particular string starts with "https://" and ends with "/MyTest".
Well, you could use regex for that. But it's clearer (and probably quicker) to just say what you mean:
str.StartsWith("https://") && str.EndsWith("/MyTest")
You then don't have to worry about whether any of the characters in your match strings need escaping in regex. (For this example, they don't.)

In .NET:
^https://.*/MyTest$

Try Expresso, good for building .NET regexes and teaching you the syntax at the same time.

HAndy tool for genrating regular expressions
http://txt2re.com/