Using JSON data retrieved with AJAX outside success function

Question

I have a problem with storing JSON that I get with AJAX, to an outside variable for further usage. Ive checked this answer (load json into variable) which is really basic, but I'm doing wrong something else. My code is below.

function showZone() {
var data=null;
$.ajax({
            url: 'http://localhost/gui/templates/tracking/show_zones.php',
            //data: 'userid='+ uid ,
            contentType: "application/x-www-form-urlencoded; charset=utf-8",
            dataType: "json",
            type: "POST",
            success: function(json) {
                data=json;
                $( '#res1' ).html( data[0].swlat );  

            }
 }); 
 return data;

}

function showZones() {
    var data=showZone();
    $( '#res2' ).html( data[0].swlat );  
}

For clearer picture of my problem I have two divs (#res1 & #res2), where I print the data. In #res1 I get the result as wanted, but #res2 doesnt print anything and I get an error 'Uncaught TypeError: Cannot read property '0' of null'. So the data is returned before ajax stores it in a variable. Is this the problem, or should I be storing json to a variable differently? Any help appreciated :)

loaded data stored in success function – Özgür Ersil May 10 '16 at 08:01 — Özgür Ersil, May 10 '16 at 08:01

score 6 · Accepted Answer · edited May 23 '17 at 11:55

You can use callback(). Consider following snippet:

function showZone() {
    var data = null;
    $.ajax({
        url: 'http://localhost/gui/templates/tracking/show_zones.php',
        //data: 'userid='+ uid ,
        contentType: "application/x-www-form-urlencoded; charset=utf-8",
        dataType: "json",
        type: "POST",
        success: function(json) {
            data = json;
            showZones(data);//callback

        }
    });
    //return data; No need to return data when we have async ajax

}
showZone(); // call when you want to make ajax call
function showZones(data) { // This function will call after ajax response
    $('#res2').html(data[0].swlat);
}

score 5 · Answer 2 · edited May 23 '17 at 10:29

5

$.ajax returns immediately return data which is executed before the function you passed as success callback was even called.So its return as undefined .Its means you can't return ajax data .

For more example How do I return the response from an asynchronous call?

But why can't you use simply like

  success: function(json) {
            data=json;
            $( '#res1' ).html( data[0].swlat );  
            $( '#res2' ).html( data[0].swlat ); 
        }

edited May 23 '17 at 10:29

Community

1
1

answered May 10 '16 at 09:09

Jack jdeoel

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Thank you for that link, I was thinking the problem is something like this yeah. I don't really need to show data in div, this was just for demonstration of my problem, I only wanted to use JSON data outside ajax function. However I'll solve this with a callback. – Gregor Stopar May 10 '16 at 09:15

Using JSON data retrieved with AJAX outside success function

2 Answers2