Regex to find last token on a string

I was wondering if there is a way having this

var string = "foo::bar"

To get the last part of the string: "bar" using just regex.

I was trying to do开发者_如何学编程 look-aheads but couldn't master them enough to do this.

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UPDATE

Perhaps some examples will make the question clearer.

var st1 = "foo::bar::0"
match should be 0

var st2 = "foo::bar::0-3aab"
match should be 0-3aab

var st3 = "foo"
no match should be found

You can use a negative lookahead:

/::(?!.*::)(.*)$/

The result will then be in the capture.

Another approach:

/^.*::(.*)$/

This should work because the .* matches greedily, so the :: will match the last occurence of that string.

Simply,

/::(.+)$/

You can't use lookaheads unless you know exactly how long a string you're trying to match. Fortunately, this isn't an issue, because you're only looking at the end of the string $.

I wouldn't use regular expressions for this (although you certainly can); I'd split the string on ::, since that's conceptually what you want to do.

function lastToken(str) {
  var xs = str.split('::');
  return xs.length > 1 ? xs.pop() : null;
}

If you really want just a regular expression, you can use /::((?:[^:]|:(?!:))*)$/. First, it matches a literal ::. Then, we use parentheses to put the desired thing in capturing group 1. The desired thing is one or more copies of a (?:...)-bracketed string; this bracketing groups without capturing. We then look for either [^:], a non-colon character, or :(?!:), a colon followed by a non-colon. The (?!...) is a negative lookahead, which matches only if the next token doesn't match the contained pattern. Since JavaScript doesn't support negative lookbehinds, I can't see a good way to avoid capturing the :: as well, but you can wrap this in a function:

function lastTokenRegex(str) {
  var m = str.match(/::((?:[^:]|:(?!:))*)$/);
  return m && m[1];
}

var string2 = string.replace(/.*::/, "");

though perhaps string isn't the best choice of name for your string?