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up vote 3 down vote favorite

Using $ to match end-of-input gives a zero-length match everywhere else but no evidence of a match with WebKit:

function showBug() {
  Result = "the end.".replace( /(end\.)([\s]|$)?/img, makeChange );
  return;
  }
function makeChange() {
  for ( var i = 0; i < arguments.length; i += 1 ) {
    document.write( "arg" + i + " -->" + arguments[ i ] + "<--"  + "<BR>" );
    }
  }

gives

arg0 -->end.<--
arg1 -->end.<--
arg2 -->undefined<--
arg3 -->4<--
arg4 -->the end.<--

for AppleWebKit/534.55.3 (KHTML, like Gecko) Version/5.1.5 Safari/534.55.3, also for AppleWebKit/535.19 (KHTML, like Gecko) Chrome/18.0.1025.162 Safari/535.19.

Opera (Presto/2.10.229 Version/11.62), FF (Gecko/20100101 Firefox/10.0.2) and IE (MSIE 8.0; Trident/4.0) all give

arg0 -->end.<--
arg1 -->end.<--
arg2 --><--
arg3 -->4<--
arg4 -->the end.<--

which means I can detect the match in $2 (it's actually about interpreting a trailing dot on a url as not being part of the url). I'm currently adding a trailing space for WebKit, and taking it off afterwards, but I'm wondering if anyone has a better solution and can confirm I should raise this as a bug.

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What happens if you remove the "?" from the regular expression? – John Fisher Apr 27 '12 at 17:11
2  
First, you don't need the | in that expression. So you could write it /(end\.)([\s])?$/ to match something before end-of-line. Second, are you trying to match 0 or 1 of 's' and '\' or are you trying to match whitespace \s (no brackets)? jsfiddle.net/2hZdT – Mathletics Apr 27 '12 at 17:13
I just tested here in Chrome and it works. – Rodrigo Manguinho Apr 27 '12 at 17:15
var t = "the end.".replace( /(end\.)([\s]|$)+/img, "###"); the value of "t" is "the ###" – Rodrigo Manguinho Apr 27 '12 at 17:16
@JohnFisher Good idea, that makes WebKit show a match occurred. But now the regex doesn't do what I want i.e. find a dot with at least one space after else eoi. – anweald Apr 27 '12 at 17:23
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1 Answer

up vote 0 down vote

If you want to match the empty string, you should include the question mark inside the capturing group, e.g.

/(end\.)((?:\s|$)?)/

The reason is that a group that does not capture anything is not the same thing that a group that captures an empty string. For example, think at what the second parentheses should capture in the following example :

/a(b(x))?c/.exec('ac');
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