Calling ajax returned outside the function?

Question

I am trying to call the ajax returned outside its function, below is the code I tried but didn't work.

var names;
$.ajax({
    type: 'POST',
    url: 'ajax/get_upcoming.php'
}).success(function (data) {
    names = data[1];
});

alert(names);

It didn't work is too generic.... But there could be an obvious reason which is asked most frequently here — PSL, Oct 05 '13 at 05:25

Vijay · Accepted Answer · 2013-10-05T05:52:02.153

0

Because success function is callback function and called when ajax request get the response , so instead try this

var names;
$.ajax({
    type: 'POST',
    url: 'ajax/get_upcoming.php',
    success: performOperation
});



function performOperation(data)
{
    // write your code here 
    names = data[1];
}

edited Oct 05 '13 at 05:52

answered Oct 05 '13 at 05:30

Vijay

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I know how to do it, but i want to use names on other function. That's the whole idea – user2310422 Oct 05 '13 at 05:32
If you want to perform some opertation or activty after ajax request you need to put it in success function only . Because when request is going to complete,nobody knows , so its better practice to create a function and you write what you want to do and call it in success function – Vijay Oct 05 '13 at 05:35
Nope, you are not helping. – user2310422 Oct 05 '13 at 05:46
Now you are helping instead of suggesting. – user2310422 Oct 05 '13 at 05:58

score 0 · Answer 2 · answered Oct 05 '13 at 05:32

0

I think alert() executes before the response comes back from the server. put console.log() and remove the alert and see the order of executions.

answered Oct 05 '13 at 05:32

Evren Bingøl

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Calling ajax returned outside the function?

2 Answers2