Return result from nested asynchronous ajax call

Question

I need to return the data from an nested ajax call but also need the ajax call to stay asynchronous.

I found some related questions but can't get it to work in the situation below. I assume returnData is being returned before the ajax call is done, but I can't find an solution.

function makeRequest(command, postData){

    var returnData;

    $.ajax({

        url: 'call.php?command='+command,
        data: postData,
        type: 'POST'

    }).done(function(data){
        returnData = data;
    });

    return returnData;
}

put the return inside the .done callback.

Aidanc
– Aidanc

2012-04-17 10:23:18 +00:00
Commented Apr 17, 2012 at 10:23 — Aidanc
– Aidanc, Commented Apr 17, 2012 at 10:23
@Aidanc — which will return the data to … where?

Quentin
– Quentin

2012-04-17 10:25:59 +00:00
Commented Apr 17, 2012 at 10:25 — Quentin
– Quentin, Commented Apr 17, 2012 at 10:25

Nicola Peluchetti · Accepted Answer · 2012-04-17 10:24:28Z

4

Yes since this call is async returnData is returned immediately. If you need to use returndata pass it to a function in the callback

function makeRequest(command, postData, functionToCallAfterAjax){

    var returnData;

    $.ajax({

        url: 'call.php?command='+command,
        data: postData,
        type: 'POST'

    }).done(function(data){
        functionToCallAfterAjax(data);
    });


}

Of course you could pass the function to call as a parameter.

This means that if your code was meant to do:

var returnedData = makeRequest(command, postData);
anotherFunction(returnedData);

you should do simply (using the code above)

makeRequest(command, postData, anotherFunction);

and everything will work

answered Apr 17, 2012 at 10:24

Nicola Peluchetti

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Comments

Quentin · Accepted Answer · 2012-04-17 10:23:48Z

0

You can't. Asynchronous events don't work like that.

The HTTP request is sent, then the function that triggers it continues immediately. The callback won't be fired until the HTTP response has arrived, but which time the calling function will have ended.

You must have your callback (the function you are passing to done()) perform whatever further processing is needed.

answered Apr 17, 2012 at 10:23

Quentin

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Comments

James Hill · Accepted Answer · 2012-04-17 10:24:04Z

0

You're returning the value of returnData before the .done() function has run. Correct your code by passing the received data to a function for processing:

function makeRequest(command, postData){
    $.ajax({
        url: 'call.php?command='+command,
        data: postData,
        type: 'POST'
    }).done(function(data){
        HandleCallBack(data);
    });
}

answered Apr 17, 2012 at 10:24

James Hill

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