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 :)
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 :)
Share Improve this question edited May 23, 2017 at 12:03 CommunityBot 11 silver badge asked May 10, 2016 at 7:57 Gregor StoparGregor Stopar 1811 silver badge11 bronze badges 2- Use callback simple. – Manwal Commented May 10, 2016 at 7:59
- loaded data stored in success function – Özgür Ersil Commented May 10, 2016 at 8:01
2 Answers
Reset to default 6You 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);
}
$.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 );
}