Passing array data

Question

Page 1 (index.php)

<?php for($i=0;$i<2;$i++) { ?>
<select id="name[<?php echo $i;?>]">
<option value="John">John</option>
<option value="Alice">Alice</option>
</select>

<input type="text" id="name-id">

<input type="text" id="location">

<?php } ?>
</table>
<script src="http://code.jquery.com/jquery-1.8.0.min.js"></script>
<script src="js/function.js"></script>

Page 2 (function.js)

for (i=0;i<2;i++){
$('select#name['+i+']').on('change',function()
{ alert(1); } ); }

The Code is not working after i add in the array.. Someone please help me..

Your are generating invalid markup, IDs are supposed to be unique. — Ram
– Ram, Commented Jan 21, 2014 at 2:18
How to do if i want to add another 5 same drop down menu in the same page? if without array — Billy
– Billy, Commented Jan 21, 2014 at 2:25
in the name is where you add the bracket to build an array <input name="some_name[]" > — Emilio Gort
– Emilio Gort, Commented Jan 21, 2014 at 2:27

xdazz · Accepted Answer · 2014-01-21 02:21:54Z

1

You have to escape the [] in the id, but rather do that, you'd better not introduce [] in id, the [] should be used in name attribute in common usage.

html:

<select id="name_<?php echo $i;?>" name="name[]">

js:

$('select#name_'+i).on('change',function() { //...

Or you could add a class and use class selector instead of id selector.

answered Jan 21, 2014 at 2:21

xdazz

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Comments

Arun P Johny · Accepted Answer · 2014-01-21 02:18:15Z

0

You need to escape the [] in id

for (var i = 0; i < 2; i++) {
    $('#name\\[' + i + '\\]').on('change', function () {
        alert(1);
    });
}

also since id should be unique in a page there is no need for the element selector select

But I would recommend adding a class to select the similar select elements instead of using id

<select id="name[<?php echo $i;?>]" class="name">
    <option value="John">John</option>
    <option value="Alice">Alice</option>
</select>

then

$('select.name').on('change', function () {
    alert(1);
});

answered Jan 21, 2014 at 2:18

Arun P Johny

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2 Comments

Billy Over a year ago

is working...Thanks.. var name=new Array(); name[i]= $('select#name\[' + i + '\]').val(); is it correct for this?

Arun P Johny Over a year ago

@user2892997 name[i]= $('select#name\\[' + i + '\\]').val(); - you need to use \` not `

Tushar Gupta - curioustushar · Accepted Answer · 2014-01-21 02:18:55Z

0

You need to escape [ AND ] with \\ in front.

$('#name\\[' + i + '\\]').on('change', function () {
    alert(1);
});

answered Jan 21, 2014 at 2:18

Tushar Gupta - curioustushar

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Comments

Stepan Yudin · Accepted Answer · 2014-01-21 02:24:58Z

0

I think that is a bad idea to make identifiers look like an array. Maybe you can use 'data' attribute?

I edited your code a little:

HTML

<?php for($i=0;$i<2;$i++) { ?>
<select id="name" data-index="[<?php echo $i;?>]">
    <option value="John">John</option>
    <option value="Alice">Alice</option>
</select>

<input type="text" id="name-id">

<input type="text" id="location">

<?php } ?>

JS

    $('body').on('change',"select#name", function () {
        alert($(this).data("index"));
    });

answered Jan 21, 2014 at 2:24

Stepan Yudin

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