how do you pass an array index to a jquery selector?

Question

How would you use an array index as a jquery selector?

for example if I have:

 array[0] = someId;
 array[1] = someId;

 for(var i; i<array.length; i++){
     $("#"+array[i]").find(.....)
 }

I would like to be able to use the value stored in the array index as my id for the selector.

thanks

what does not work with your code (except the one quotation mark to much)? — Bergi
– Bergi, Commented Mar 15, 2012 at 22:18
$("#"+array[i]).find("#someELement"); - You had a stray closing " in your string, no need for it. — Ohgodwhy
– Ohgodwhy, Commented Mar 15, 2012 at 22:18
Closing and down-voting because this is a syntax error. Learn your tools. — user166390
– user166390, Commented Mar 15, 2012 at 22:47

SLaks · Accepted Answer · 2012-03-15 22:18:33Z

2

Your code works fine, except for the extra ".
jQuery selectors are ordinary strings.

answered Mar 15, 2012 at 22:18

SLaks

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Jasper · Accepted Answer · 2012-03-15 22:25:25Z

2

Remove the last quote in selector array[i] will be a string and all you are doing is concatenating a string to create the selector

$("#"+array[i] ).find(.....)

edited Mar 15, 2012 at 22:25

Jasper

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answered Mar 15, 2012 at 22:18

charlietfl

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Comments

osahyoun · Accepted Answer · 2012-03-15 22:21:21Z

1

Don't forget to set i to 0:

 array[0] = someId;
 array[1] = someId;

 for(var i=0; i<array.length; i++){
     $("#"+array[i]).find(.....)
 }

answered Mar 15, 2012 at 22:21

osahyoun

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Comments

Jasper · Accepted Answer · 2012-03-15 22:23:08Z

1

You can create a multiple selector by joining the indexes of your array:

$('#' + array.join(', #')).find(.....)

This way you don't loop through anything, you let jQuery deal with the fact that you're doing something to multiple elements.

Here is a demo: http://jsfiddle.net/y7UQ9/

Docs for multiple selectors: http://api.jquery.com/multiple-selector/

answered Mar 15, 2012 at 22:23

Jasper

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Comments

Ben Everard · Accepted Answer · 2012-03-15 22:19:02Z

0

You've got a syntax error to start with:

$("#" + array[i]).find(.....)

answered Mar 15, 2012 at 22:19

Ben Everard

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