I would expect this line of JavaScript:
\"foo bar baz\".match(/^(\\s*\\w+)+$/)
to return something like:
[\"foo bar baz\",
When you repeat a capturing group, in most flavors, only the last capture is kept; any previous capture is overwritten. In some flavor, e.g. .NET, you can get all intermediate captures, but this is not the case with Javascript.
That is, in Javascript, if you have a pattern with N capturing groups, you can only capture exactly N strings per match, even if some of those groups were repeated.
So generally speaking, depending on what you need to do:
/(pattern)+/
, maybe match /pattern/g
, perhaps in an exec
loop
Here's an example of matching <some;words;here>
in a text, using an exec
loop, and then splitting on ;
to get individual words (see also on ideone.com):
var text = "a;b;<c;d;e;f>;g;h;i;<no no no>;j;k;<xx;yy;zz>";
var r = /<(\w+(;\w+)*)>/g;
var match;
while ((match = r.exec(text)) != null) {
print(match[1].split(";"));
}
// c,d,e,f
// xx,yy,zz
The pattern used is:
_2__
/ \
<(\w+(;\w+)*)>
\__________/
1
This matches <word>
, <word;another>
, <word;another;please>
, etc. Group 2 is repeated to capture any number of words, but it can only keep the last capture. The entire list of words is captured by group 1; this string is then split
on the semicolon delimiter.
Unless you have a more complicated requirement for how you're splitting your strings, you can split them, and then return the initial string with them:
var data = "foo bar baz";
var pieces = data.split(' ');
pieces.unshift(data);
try using 'g':
"foo bar baz".match(/\w+/g)
How's about this? "foo bar baz".match(/(\w+)+/g)