I have two variables defined like this:
var $a = $('#a'),
$b = $('#b');
How can I rewrite the following line using $a and $b?
$('#a, #b').click(function(e){...});
views:
41answers:
3Wouldn't it be better to just use $('#a, #b').click() along with $a, $b then?
Vincent
2010-10-19 09:03:33
Absolutely, much better, but I thought you don't want to do that and ask how to do it assuming you already have the $a and $b variables.
Darin Dimitrov
2010-10-19 09:05:03
Keep in mind this only works on $a and $b's first element. If you change $a or $b to a selector that returns more then one element only the first will be bound. But I agree with Darin $('#a, #b') is a better solution if you only use them once.
fredrik
2010-10-19 09:25:44
A:
Depends of what you want to do next with your vars.
You could just define them as one:
var $a = $('#a, #b');
$a.click()
Or use them separately:
/* This way the click event is applied even if each selector returns
more then one element. And $a and $b is preserved */
$([].concat($a.toArray()).concat($b.toArray())).click(function () {
console.log(this)
});
EDIT
Updated code.
..fredrik
fredrik
2010-10-19 08:57:35
No, they're not equivalent. The second will make `$b` the context.
Matthew Flaschen
2010-10-19 08:58:55
I did a test case: var a = $('div'), b = $('span'); $(a,b).click(function { console.log(this) }); console only returns div's.
fredrik
2010-10-19 09:22:19
@fred, your example isn't complete. But read [the docs](http://api.jquery.com/jQuery/#jQuery1) on the main function, and it's clear which parameter is which.
Matthew Flaschen
2010-10-19 09:27:24
+1
A:
$a.add($b).click(function(e){...});
add returns a new node set holding the union. $b can be "pretty much anything that $() accepts."
Matthew Flaschen
2010-10-19 09:08:45