I am having trouble matching a specific pattern in a backgroundImage css expression. I want to get the u/user/rest/of/stuff out of "url(foo/bar/u/user/rest/of/stuff)". However I am not sure how to get only the second u (the u in url would be the first), plus everything after that up to the second parenthesis.
2 Answers
Use word boundary(\b) ,
> "url(foo/bar/u/user/rest/of/stuff)".match(/\bu\b[^)]*/g);
[ 'u/user/rest/of/stuff' ]
OR
Use a look-ahead (?=...)
> "url(foo/bar/u/user/rest/of/stuff)".match(/u[^()]*(?=\))/g);
[ 'u/user/rest/of/stuff' ]
Comments
For I produce a lot of typos in regex and for better readability I use functions whenever possible for this kind of stuff...
function GetSubstring(str, from, to) {
return str.substring(str.lastIndexOf(from)+from.length,str.lastIndexOf(to));
}
And then use
GetSubstring('url(foo/bar/u/user/rest/of/stuff)','url(foo/bar/',')');
url()value first, then isolate the string inside it and work with that.