jquery selector performance

Question

Is there a more performant way to write this.

$('#test').find('option:selected[value!=""]')

not at all, Just want to see if there are better ways to writing this. — Hussein, Feb 08 '11 at 08:47

jAndy · Accepted Answer · 2011-02-09T07:05:14.783

5

You can tweak it at little, but using methods instead of Sizzle:

$('#test').find('option').filter(function() {
    return this.selected && this.value.length
});

Benchmark: http://jsperf.com/sizzle-vs-methods-filter/12

.filter() is about 70% faster for me.

edited Feb 09 '11 at 07:05

answered Feb 08 '11 at 08:36

jAndy

I tried this.selected && this.value.length, Console.log is giving jquery() and not a value. – Hussein Feb 09 '11 at 03:45
@alex: whoops, I forgot the `return statement`, see update. Sorry. – jAndy Feb 09 '11 at 06:42
@jAndy, can you re-run the benchmark with the return statement included? I'd be interested to know if it affects the results. – a'r Feb 09 '11 at 06:52
@ar: I updated the benchmark aswell, does not affect the performance really. – jAndy Feb 09 '11 at 07:06
@jandy, Thanks, i figured the return was missing after my post. – Hussein Feb 09 '11 at 18:23
Wow. I never would have guessed that .filter() vs. selector could make that big of a difference. If anything, I thought it'd be slower. – Andrew Ensley Dec 11 '12 at 22:24

score 0 · Answer 2 · answered Feb 08 '11 at 08:37

0

Well, there will always only be one selected, so you don't need a find() handler in my opinion.

I'll just write it like this:

$('#test option:selected[value!=""]')

I haven't tested it yet.

answered Feb 08 '11 at 08:37

Anriëtte Myburgh

Using http://jsperf.com this is giving a 2% slower result then find. using filter is the way to go. – Hussein Feb 08 '11 at 20:29

2 Answers2