Jquery array.push() not working

Question

I have been trying to add variables from a dropdown into an array using Jquery array.push() function but for some odd reason it is not working. Following is the jsFiddle link: http://jsfiddle.net/dKWnb/3/

JavaScript :

$("#test").live("click",function() {
       var myarray = new Array();
        myarray.push($("#drop").val());
        alert(myarray);
});

HTML

<Select name=drop id=drop>
    <option value=1>1</option>
    <option value=2>2</option>
</select>
<input type=button name=test id=test value=test>

What should it do? Because it does here exactly what it looks it should do. — netiul
– netiul, Commented Dec 7, 2011 at 10:09
Just a note: always declare arrays this way var myarray = []; — Didier Ghys
– Didier Ghys, Commented Dec 7, 2011 at 10:09
@Matt. Long story short, everything can be re-defined in javascript, so is Array(). Using [] guaranties you'll have an array. This is also more efficient. Check this answer that explains it pretty clearly. — Didier Ghys
– Didier Ghys, Commented Dec 7, 2011 at 10:18
For the record, the .push() method is native JavaScript and got nothing to do with jQuery. — user447356
– user447356, Commented Dec 7, 2011 at 10:22

Manse · Accepted Answer · 2011-12-07 10:42:28Z

33

~~Your HTML should include quotes for attributes : http://jsfiddle.net/dKWnb/4/~~

Not required when using a HTML5 doctype - thanks @bazmegakapa

You create the array each time and add a value to it ... its working as expected ?

Moving the array outside of the live() function works fine :

var myarray = []; // more efficient than new Array()
$("#test").live("click",function() {
        myarray.push($("#drop").val());
        alert(myarray);
});

http://jsfiddle.net/dKWnb/5/

Also note that in later versions of jQuery v1.7 -> the live() method is deprecated and replaced by the on() method.

edited Dec 7, 2011 at 10:42

answered Dec 7, 2011 at 10:09

Manse

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2 Comments

kapa Over a year ago

He uses an HTML5 doctype. It is not obligatory to use quotes around attribute values.

Manse Over a year ago

@bazmegakapa ok understood - i have learnt something there - answer updated

Interrobang · Accepted Answer · 2011-12-07 10:09:00Z

7

Your code alerts the current value of the dropdown for me, showing that it has properly pushed into the array.

Are you wanting to keep old values and append? You're recreating the array each time, meaning that the old value gets clobbered.

Here's some updated code:

var myarray = [];
$("#test").click(function() {
    myarray.push($("#drop").val());
    alert(myarray);
});

jsFiddle

answered Dec 7, 2011 at 10:09

Interrobang

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Comments

sifr_dot_in · Accepted Answer · 2019-06-18 05:33:40Z

4

another workaround:

var myarray = [];
$("#test").click(function() {
    myarray[index]=$("#drop").val();
    alert(myarray);
});

i wanted to add all checked checkbox to array. so example, if .each is used:

var vpp = [];
var incr=0;
$('.prsn').each(function(idx) {
   if (this.checked) {
       var p=$('.pp').eq(idx).val();
       vpp[incr]=(p);
       incr++;
   }
});
//do what ever with vpp array;

answered Jun 18, 2019 at 5:33

sifr_dot_in

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Comments

Dhruv Hirpara · Accepted Answer · 2021-12-30 10:17:54Z

-1

var array = [];
var element = "anything you want in the array";
array.push(element); // array = [ "anything you want in the array" ]

answered Dec 30, 2021 at 10:17

Dhruv Hirpara

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1 Comment

Community Over a year ago

As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.

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