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Regex for finding the text inside parentheses followed by #en : "example"#en [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

Regex: Match a string between two tags in a string

I am new to Regexp. I am struck in writing regexp for below scenario. Can some one please help me in solving this?
If i have a String like the following:
<Tag1 attr="test"/>
<Tag2>
<Tag4 attr="test"/>
<Tag5 attr="test"/>
</Tag2>
<Tag3 attr="test"/>
Whats the regex to match 'test' between the <Tag2> and </Tag2> tags?
Output should match 'test' in both Tag4 and Tag5...
Any help would be highly appreciated..

Why are you using a regex for this? I am not familiar with the Java libraries, but I would imagine there is a library that would allow you to do XQueries using XPaths. That would be the simpler approach.
Here is a website that shows examples
Here is a SO question on XPath in Java

XPath is really more appropriate for this. This looks like duplicate post. Original
Perl has a couple of good xpath parsers on CPAN. But here's a good page on multiline regex parsing if you absolutely must use it.

All said before is totally true - however if you still want to practice some regex heres an alternative:
Doing it in one match is not possible since one of the inner groups will always be discarded (see this) , so you'll have to extract the inner passage first.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexTagParse {
static String html = "<Tag1 attr=\"test\"/><Tag2> <Tag4 attr=\"test_one\"/> <Tag5 attr=\"test_two\"/></Tag2><Tag3 attr=\"test\"/>";
public static void main(String[] args) {
Matcher mat1 = Pattern.compile("Tag2>(.*)</Tag2").matcher(html);
mat1.find();
Matcher mat2 = Pattern.compile("<[^<>]*attr=\"([^\"]+)\"[^<>]>").matcher(mat1.group(1));
while(mat2.find()){
System.out.println(mat2.group(1));
}
}
}
anyways, you'd be much better off using XPath :)

I'm not in practice with java, but I can offer some guidance to the regular expression, I hope. If you know what the specific attribute and value is that you're looking for, you can use something like the following:
Pattern pattern = Pattern.compile("<tag[45].*attr\s*=\s*[\"']test['\"][^>]*>", Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher("<Tag1 attr='test'/><Tag2><Tag4 attr='test'/><Tag5 attr='test'/></Tag2><Tag3 attr='test'/>");
matcher.matches();
the regex is made up of the following components:
match the literal string:
followed by either a 4 or a 5 (the [45] designation)
followed by any number of characters preceding the literal string: attr
followed by any number of spaces
followed by the literal character: =
followed by any number of spaces
followed by either the ' or " character
followed by the string literal: test
followed by either the ' or " character
followed by any character that is not >
followed by >
the point in adding some of these extra bits is simply to highlight that you may need/want to consider accounting for different coding styles, etc. note: I took the easy away out by setting the pattern as case-insensitive, but you can omit that and change your expression to check for the appropriate case (for example, if your attribute value is case-sensitive, you can change the 'tag' literal to be [tT][aA][gG] in order to allow matching the tag to be case-insensitive.
I'm apparently too slow to type, since jvataman has already answered your question, but perhaps there is some value in my writeup, so I'll post anyway.

Java regex matcher.find fails occasionally

I have regexp which parses all names of used freemarker macros in template (for example from <#macroName /> I need only macroName). Templates are usually quite large (round 30 thousand characters).
Java code with regex looks like:
Pattern pattern = Pattern.compile(".*?<#(.*?)[ /].*?",
Pattern.DOTALL | Pattern.UNIX_LINES);
Matcher matcher = pattern.matcher(inputText);
while(matcher.find()){
//... some code
}
But sometimes happens that I get this exception:
java.util.regex.Pattern$Curly.match1(Pattern.java:3814)
java.util.regex.Pattern$Curly.match(Pattern.java:3763)
java.util.regex.Pattern$Start.match(Pattern.java:3072)
java.util.regex.Matcher.search(Matcher.java:1116)
java.util.regex.Matcher.find(Matcher.java:552)
...
Does anybody know why it happens or could anybody make me sure if the regexp I'm using is optimized well?
thank you

For <#macro macroName /> your regex looks a little bit convoluted. Either there are things (special cases) that <#macro macroName /> don't describe, or the regex is trying too hard. Try:
<#macro\s+(\S+)\s+/>
You should have now the macro's name in group #1.

You can get rid of the leading .*? because you don't need to consume the text before/between the matches. The regex engine will take care of scanning for the next match, and it will do it a lot more efficiently than what you're doing. Just give it the pattern for the tag itself and get out of its way.
You can get rid of the trailing .*? because it never does anything. Think about it: it's trying to match zero or more of any characters, reluctantly. That means the first thing it tries to do is match nothing. That attempt will succeed (it's always possible to match nothing), so it never tries to consume more characters.
You probably want something like this ():
<#(\w+)[\s/]
...or in Java-speak:
Pattern p= Pattern.compile("<#(\\w+)[ /]");
You don't need DOTALL (no dots) or any other modifiers.

Possible Regular Expression Question

I have a simple program that looks up details of an IP you give it, and I will show you an example of some of my code
int regIndex = src.indexOf("Region:") + 16;
int endIndex = src.indexOf("<", regIndex);
String region = src.substring(regIndex, endIndex);
if(regIndex == 15) region = "None";
int counIndex = src.indexOf("Country:") + 17;
int couneIndex = src.indexOf(" <", counIndex);
String country = src.substring(counIndex, couneIndex);
As you can see, it is definitely not the most efficient way to do this. The website I am using gives the information like this: http://whatismyipaddress.com/ip/1.1.1.1
I have never really used Regular Expressions before, but it seems to me like there might be one that could really make this more efficient and easier to program, but I've been looking around and I'm pretty lost.
So basically my question is, how could I use a Regular Expression for this (Or if there is another more efficient way).
Any help would be great,
Thanks :)

You can do something like this:
String s = "bla Country: Australia <bla";
Pattern pattern = Pattern.compile("Country: (.*) [<]");
Matcher matcher = pattern.matcher(s);
if(matcher.find()) {
System.out.println("Country = " + matcher.group(1));
}

The source would look like this
<tr><th>Country:</th><td>Australia <img src="http://whatismyipaddress.com/images/flags/au.png" alt="au flag"> </td></tr>
To use regular expression means to match a pattern.
The pattern that indicates your wanted data is pretty straight forward Country:. You need also to match the following tags like <\/th><td>. The only thing is you need to escape the forward slash. Then there is the data you are looking for, I would suggest to match everything that is not a <, so [^<], this is a capturing group with a negation at the beginning, meaning any character that is not a <, to repeat this add a + at the end, meaning at least one of the preceding character.
So, the complete thing should look like this:
Country:<\/th><td>\s*([^<]+)\s*<
I added here also the brackets, they mean put the found pattern into a variable, so your result can be found in capturing group 1. I added also \s*, this is a whitespace character repeated 0 or more times, this is to match whitespace before or after your data, I assume that you don't need that.

Firstly there are some online sites that can help you to develop a regular expression. They let you enter some text, and a regular expression and then show you the result of applying the expression to the text. This saves you having to write code as you develop the expression and expand your understanding. A good site I use alot is FileFormat regex because it allows me to test one expression against multiple test strings. A quick search also brought up regex Planet, RegExr and RegexPal. There are lots of others.
In terms of resources, the Java Pattern class reference is useful for Java development and I quite like regular-expression.info as well.
For your problem I used fileFormat.info and came up with this regex to match "http://whatismyipaddress.com/ip/1.1.1.1":
.*//([.\w]+)/.*/(\d+(?:.\d+){3})
or as a java string:
".*//([.\\w]+)/.*/(\\d+(?:.\\d+){3})"
A quick break down says anything (.*), followed by two slashes (//), followed by at least one or more decimal points or characters (([.\w]+)), followed by a slash, any number of characters and another slash (/.*/), followed by at least 1 digit ((\d+), followed by 3 sets of a decimal point and at least one digit ((?:.\d+){3})). The sets of brackets around the server name part and the IP part are called capturing groups and you can use methods on the Java Matcher class to return the contents of these sections. The ?: on the second part of the ip address tells it that we are using the brackets to group the characters but it's not to be treated as a capturing group.
This regex is not as strict or as flexible as it should be, but it's a starting point.
All of this can be researched on the above links.

How do I make this regex more general, sometimes it works and sometimes it doesn't

I have the following regex that I am using in a java application. Sometimes it works correctly and sometimes it doesn't.
<!-- <editable name=(\".*\")?> -->(.*)<!-- </editable> -->
Sometimes I will have whitespace before/after it, sometimes there will be text. The same goes for the region within the tags.
The main problem is that name=(\".*\")?> sometimes matches more than it is supposed to. I am not sure if that is something that is obvious to solve, simply looking at this code.

XML is not a regular language, nor is HTML or any other language with "nesting" constructs. Don't try to parse it with regular expressions.
Choose an XML parser.

As others have pointed out, the greedy .* (dot-star) that matches the "name" attribute needs to be made non-greedy (.*?) or even better, replaced with a negated character class ([^"]*) so it can't match beyond the closing quotation mark no matter what happens in the rest of the regex. Once you've fixed that, you'll probably find you have the same problem with the other dot-star; you need to make it non-greedy too.
Pattern p = Pattern.compile(
"<!--\\s*<editable\\s+name=\"([^\"]*)\">\\s*-->" +
"(.*?)" +
"<!--\\s*</editable>\\s*-->",
Pattern.DOTALL);
I don't get the significance of your remarks about whitespace. If it's linefeeds and/or carriage returns you're talking about, the DOTALL modifier lets the dot match those--and of course, \s matches them as well.
I wrote this in the form of a Java string literal to avoid confusion about where you need backslashes and how many of them you need. In a "raw" regex, there would be only one backslash in each of the whitespace shorthands (\s*), and the quotation marks wouldn't need to be escaped ("[^"]*").

I would replace that .* with [\w-]* for example if name is an identifier of some sort.
Or [^\"]* so it doesn't capture the end double quote.
Edit:
As mentioned in other post you might consider going for a simple DOM traversal, XPath or XQuery based evaluation process instead of a plain regular expression. But note that you will still need to have regex in the filtering process because you can find the target comments only by testing their body against a regular expression (as I doubt the body is constant judjing from the sample).
Edit 2:
It might be that the leading, trailing or internal whitespaces of the comment body makes your regexp fail. Consider putting \s* in the beginning and at the end, plus \s+ before the attribute-like thing.
<!--\s*<editable\s+name=(\"[^\"]*\")?>\s*-->(.*)<!--\s*</editable>\s*-->
Or when you are filtering on XML based search:
"\\s*<editable\\s+name=(\"[^\"]*\")?>\\s*"
"\\s*</editable>\\s*"
Edit 3: Fixed the escapes twice. Thanks Alan M.

the * multiplier is "greedy" by default, meaning it matches as much as possible, while still matching the pattern successfully.
You can disable this by using *?, so try:
(\".*?\")