CODEWITHSUNDEEP
javascriptjqueryjsonajax
Gregor Stopar
Gregor Stopar

Reputation: 181

Using JSON data retrieved with AJAX outside success function

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 :)

Upvotes: 8

Views: 1453

Answers (2)

Jack jdeoel
Jack jdeoel

Reputation: 4584

$.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 ); 
        }

Upvotes: 5

Manwal
Manwal

Reputation: 23816

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);
}

Upvotes: 6

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