匹配双引号或单引号之间的文本

Question

I need some regex help. I have been banging my head on this last few hours. I need to match some strings in a minified file.

Sample string:

var a ='abc'; var b = 'http://a/that.dude.js/v1/'; var c = 'def'; var d = 'https://b/that.dude.js/v1/';
var basePath = "http://othersite/that.dude.js/v1/";

I want to match full text inside single or double quotes that contains that.dude.js/v1 . I tried:

/('|").+that.dude.js\/v1\/('|")/g

...but this matches the full line when there are multiple occurrences in the same line.

My expected match will be:

http://a/that.dude.js/v1/
https://b/that.dude.js/v1/
http://othersite/that.dude.js/v1/

Here is what I have tried: http://regexr.com/3cv62

Answer 1

Try this one:

/(["'])[^"']+that\.dude\.js\/v1\/\1/g

The only modification was to change . to [^"'] this doesn't allow quotes between the quotes.

Answer 2

If you have single quotes inside double quoted strings, you need to capture the quote delimiter and use a backreference to match exactly the same trailing delimiter:

(['"])([^"'\s]*that\.dude\.js\/v1[^"'\s]*)\1

See the regex demo .

Since you have URLs, you can safely match them with [^"'\\s]* (one or more symbols other than " , ' and whitespace). The regex matches:

• (['"]) - the leading quote delimiter (captured into Group 1 so that we could match the same trailing delimiter)
• ([^"'\\s]*that\\.dude\\.js\\/v1[^"'\\s]*) - Group 2 matching
  • [^"'\\s]* - 0+ symbols other than " , ' and whitespace
  • that\\.dude\\.js\\/v1 - that.dude.js/v1
  • [^"'\\s]* - ibid.
• \\1 - trailing delimiter that is the same as the leading one

The result will be in Group 2:

 var re = /(['"])([^"'\\s]*that\\.dude\\.js\\/v1[^"'\\s]*)\\1/g; var str = 'var a =\\'abc\\'; var b = \\'http://a/that.dude.js/v1/\\'; var c = \\'def\\'; var d = \\'https://b/that.dude.js/v1/\\';\\nvar basePath = "http://othersite/that.dude.js/v1/";'; var res = []; while ((m = re.exec(str)) !== null) { res.push(m[2]); } document.body.innerHTML = "<pre>" + JSON.stringify(res, 0, 4) + "</pre>"; 

Note that to make it even more generic, you could use a tempered greedy token :

(['"])((?:(?!\1).)*that\.dude\.js\/v1(?:(?!\1).)*)\1
       ^^^^^^^^^^^^                  ^^^^^^^^^^^^  

See another demo

The (?:(?!\\1).) token will match any character(s) but a newline that are not equal to the value referred to by the \\1 backreference.