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){...});
Add a comment
|
4 Answers
$([$a.get(0), $b.get(0)]).click(function(e) { ... });
3 Comments
Vincent
Wouldn't it be better to just use $('#a, #b').click() along with $a, $b then?
Darin Dimitrov
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.
fredrik
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.
$a.add($b).click(function(e){...});
add returns a new node set holding the union. $b can be "pretty much anything that $() accepts."
Comments
It seems you ca do just this:
$(deleteBtn).add(copyBtn).add(moveBtn).click(function(){});
As it says here http://api.jquery.com/add/
Comments
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
4 Comments
Matthew Flaschen
No, they're not equivalent. The second will make
$b the context.Matthew Flaschen
No, the second parameter ($b) is the context.
fredrik
I did a test case: var a = $('div'), b = $('span'); $(a,b).click(function { console.log(this) }); console only returns div's.
Matthew Flaschen
your example isn't complete. But read the docs on the main function, and it's clear which parameter is which.