Concatenate JQuery variables

Question

I need to add a number to the end of each javascript variable and JQuery element.

Here's my code:

$("#insert1").click(function(){
    var collegeId1=$("#collegeId1").val();
     $.post('insert.php', {collegeId: collegeId1});
    return false;
});

$("#insert2").click(function(){
    var collegeId2=$("#collegeId2").val();
    $.post('insert.php', {collegeId: collegeId2});
    return false;
});

... and so on. I need to continue adding numbers to the end of each element and variable (i.e. collegeId[number] and ("#insert[number]"). I've tried looping, .append, +, etc and nothing seems to be working.

Any help would be greatly appreciated!

It is not possible in javascript like php $$var; U should use array instead variable name — Ts8060
– Ts8060, Commented Oct 3, 2013 at 4:56
it seems you are making it more complex.. post your html code. — Mr_Green
– Mr_Green, Commented Oct 3, 2013 at 5:04

Stalinko · Accepted Answer · 2013-10-03 05:00:19Z

6

Correct way to do this is using "data"-fields You have buttons like this:

<button data-id="1" class="insert">

And just one function for all this shit:

$('.insert').click(function(){
    var id = $(this).data('id');
    var collegeId = $('#collegeId' + id).val();
    $.post('insert.php', {collegeId: collegeId});
    return false;
});

Also you can store collegeId right in the button-tag (dont write "data-collegeId" since data-fields don't support uppercase):

<button data-collegeid="1" class="insert">

then you get it in 1 line:

var collegeId = $(this).data('collegeid');

answered Oct 3, 2013 at 5:00

Stalinko

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Phil · Accepted Answer · 2013-10-03 05:04:29Z

0

If you must work with the markup you've already got, try this...

$('[id^="insert"]').filter(function() {
    return /insert\d+$/.test(this.id); // filter to make sure the id matches /insert\d+/
}).on('click', function(e) {
    e.preventDefault();
    var i = /\d+$/.exec(this.id)[0], // fetch index from id
        collegeId = $('#collegeId' + i).val();
    $.post('insert.php', {collegeId: collegeId});
})';

edited Oct 3, 2013 at 5:04

answered Oct 3, 2013 at 4:57

Phil

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