Using variables and elements together in jQuery multiple selectors

Question

So...I'm trying to shrink down some JS code a bit, and I have things like...

$('#element-A'+idx).hide();
$('#element-R'+idx).hide();
$('#element-Z'+idx).hide();

Is there a reliable way I can smash these down into one statement without taking a performance hit?

I tried:

$('#element-A'+idx,'#element-R'+idx,'#element-Z'+idx).hide();

...but that just gave me an error, which actually is what I thought would happen.

I also tried:

$(['#element-A'+idx,'#element-R'+idx,'#element-Z'+idx]).hide();

...but I also got an error from that.

The only other possibility I can think of is:

$('#element-A'+idx).add('#element-R'+idx).add('#element-Z'+idx).hide();

But doesn't that create a new object, in effect taking up more memory??

What should I do?? Or am I better off just leaving them as separate commands?

SLaks · Accepted Answer · 2013-06-17 20:28:56Z

1

You need to pass a single selector string:

$('#element-A'+idx + ', #element-R'+idx + ', #element-Z'+idx).hide();

Or, better yet, add a common class to all of the elements, and just use that instead.

answered Jun 17, 2013 at 20:28

SLaks

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1 Comment

The comma inside the quotes fixed it. Thank you!

Joe · Accepted Answer · 2013-06-17 20:30:59Z

1

Your commas need to be inside the quotes:

$('#element-A'+idx,'#element-R'+idx,'#element-Z'+idx).hide();

should be

$('#element-A' + idx + ', #element-R' + idx + ', #element-Z' + idx).hide();

answered Jun 17, 2013 at 20:30

Joe

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