使用AJAX检索外部成功函数的JSON数据

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 :)

Answer 1

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);
}

Answer 2

$.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 ); 
        }