how to pop(delete) values randomly using Jquery

Question

i am having the code like

<input type="checkbox" class="selectionne" id="first" value="1">
<input type="checkbox" class="selectionne" id="second" value="2">
    <input type="checkbox" class="selectionne" id="second" value="3">
        <input type="checkbox" class="selectionne" id="second" value="4">
            <input type="checkbox" class="selectionne" id="second" value="5">
                <input type="checkbox" class="selectionne" id="second" value="6">

jquery

var elems = [];
$('.selectionne').change(function(){
  this.checked ? elems.push($(this).val()) : elems.pop($(this).val());
  alert(elems);
});

Actually am pushing values to var elems by ckecking the checkboxes.pushing happening properly, But poping is happening like how i pushed values like 1,2,3,4,5 in reverse order its poping. i dont want like that, if i check randomly also it has to pop randomly only..

check here jsfiddle its working like wat i was explained. i want to pop randomly after unchecking check boxes....

plz help me out.. thanks in advance...

Note IDs must be unique

Anton
– Anton

2014-06-13 13:03:31 +00:00
Commented Jun 13, 2014 at 13:03 — Anton
– Anton, Commented Jun 13, 2014 at 13:03

Barmar · Accepted Answer · 2014-06-13 13:08:49Z

1

pop always removes the last element, it's the opposite of push. If you want to remove a specific element by value, use:

function removeValue(array, value) {
    array.splice(array.indexOf(value), 1);
};

answered Jun 13, 2014 at 13:08

Barmar

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Comments

Muhammad Saifullah · Accepted Answer · 2014-06-13 13:44:21Z

0

Try this

var elems = [];
$('.selectionne').change(function(){
if(this.checked){
    elems.push(this.id) 
}
else {
    var toBeDeleted=this.id;  
    elems = jQuery.grep(elems, function(value) {
    return value != toBeDeleted; 
    });
    alert(elems);
}


});

http://jsfiddle.net/XBe89/

Hope this helps

answered Jun 13, 2014 at 13:44

Muhammad Saifullah

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