input an array of id's into a jquery selector

Question

So I'm trying to input an array into a jquery selector and for some reason I can't get it to work at all. I've tried a few approaches below 1st:

var id = ["#id1", "#id2", "#id3"];
$(id[0]).css('color', '#000');

my other attempt:

var id = [];
for(var x = 0;x < 3; x++){
id[x] = x;
$("#id"+id[0]).css('color', '#000');
}

So I'm wondering what might be the approach to solving this?

example of what I'm using this for:

for(var x = 0;x < 28; x++){
function randomizerlet(){
randlet[x] = Math.floor(Math.random()*156);
timesletrun[x] += 1;
if (masterlet[randlet[x]] == letter[x]){
$('#flipL4_1').css('background-position', masterlet[randlet[x]]);
clearInterval(intervallet[x]);
}
else{
    if(timesletrun[x] == 300){
        masterlet[randlet[x]] = letter[x];
        $('#flipL4_1').css('background-position', masterlet[randlet[x]]);
        clearInterval(intervallet[x]);
        }
    else{
        $('#flipL4_1').css('background-position', masterlet[randlet[x]]);
        }}


        }
var intervallet[x] = setInterval(function() {
 randomizerlet();}, 10);
}

If you are trying to do different things to id's in the list then you can't be efficient. If you are setting everything in the lists color to black then Matt's answer along with his 'use a class' suggestion is ideal. — Sanchit
– Sanchit, Commented Apr 1, 2013 at 4:23

Matt Ball · Accepted Answer · 2013-04-01 04:09:18Z

5

Join the array elements into a single string:

var id = ["#id1", "#id2", "#id3"];
$(id.join(',')).css('color', '#000');

Or use a loop:

for (var i=0; i<id.length; i++) {
    $(id[i]).css(...);
}

Or a different kind of loop:

$.each(id, function (i, elt) {
    $(elt).css(...);
});

Or an even fancier sort of loop:

var $elements = $.map(id, function (elt, i) {
    return $(elt);
});
$elements.css(...);

All that said, it would probably be better to simply use a common class instead of IDs. Any time you think you need number-like indices, that's a strong hint that you might be better off using classes instead.

edited Apr 1, 2013 at 4:09

answered Apr 1, 2013 at 4:03

Matt Ball

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5 Comments

Keleko Over a year ago

I need to have them separated, not have all of them the same.

Matt Ball Over a year ago

I do not understand. Please elaborate.

Keleko Over a year ago

the $("#id1").css('color', '#000'); is inside a function whereas $("#id2").css('color', '#000'); will be in a different function cause they are recieving different inputs for .css('color', 'different variable').

Matt Ball Over a year ago

You still haven't explained enough for me to do your thinking for you. I'm not a mind reader, so you're on your own for this one.

Keleko Over a year ago

edited posting for a better understanding of what I need, the id needs to loop through as well, each id is not the same.

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