r = /\u0020+$/g; p = r.exec( ); q = r.exec( ); JSON.stringify([p, q])
// [[\ \],null]
Why does calling exec the second time generate null? When I try the
regular expression without the /g flag, I get:
// [[\ \],[\ \]]
--
The first step in confirming there is a bug in someone else's work
expected by specs ... try this
r = /\u0020+$/g; p = r.exec( ); r.lastIndex = 0; q = r.exec( );
alert(JSON.stringify([p, q]))
and you are good to go
... without resetting lastIndex, you can also use .replace(r,
'whatever') and it's going to work every time.
Have a look at these slides to
Yes, the /g flag is confusing, because the regular expression kind of turns
into an iterator. I’d much prefer an actual iterator-based API that doesn’t
mutate the regular expression.
On Jul 16, 2014, at 9:26 , Alex Vincent ajvinc...@gmail.com wrote:
r = /\u0020+$/g; p = r.exec( ); q =
You could wrap it in an iterator :-)
http://www.es6fiddle.net/hxoczoqg/
```javascript
function* match(regex, haystack){
var {source, ignoreCase} = regex;
var newRegex = new RegExp(source, ignoreCase ? 'gi' : 'g');
var m = null;
while (m = newRegex.exec(haystack)) {
yield m;
}
}
var
you can also use this code ... like today, with every browser, and right
away ...
```javascript
function match(regex, haystack){
var result = [], g = regex.global, match;
do { match = regex.exec(haystack) }
while(match result.push(match) g);
if (!g) regex.lastIndex = 0;
return
function* match(regex, haystack){
var {source, ignoreCase} = regex;
var newRegex = new RegExp(source, ignoreCase ? 'gi' : 'g');
var m = null;
while (m = newRegex.exec(haystack)) {
yield m;
}
}
this code is not best in that case:
var regex = /\d+/;
var first = match(regex, 1 2 3 4);
6 matches
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