Javasctip variable not assigned

Question

Why is username undefined? I can be sure that the JSON file format is right.

I need set username as the global variable, is there any way i can workaround this?

var loginProfile = 'link to JSON file';
var username;
$.getJSON(loginProfile, function (data) {
    username = data.USERNAME;
});

alert(username);

please google and read about `callback` it will help :)! – Kai Dec 02 '17 at 16:43 — Kai, Dec 02 '17 at 16:43

score 0 · Answer 1 · answered Dec 02 '17 at 16:43

0

Because your getJSON is asynchronous, the value is undefined when you put alert. Instead you can use callback or function as below and display. Try this,

var loginProfile = 'link to JSON file';
var username;
$.getJSON(loginProfile, function (data) {
    username = data.USERNAME;
    printUserName(username);
});

function printUserName(name){
  alert(name);
}

answered Dec 02 '17 at 16:43

Sajeetharan

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I need it assigned to a global variable. The function wont work for my case. – Programmer Dec 02 '17 at 16:56
still you can assign to a global variable – Sajeetharan Dec 02 '17 at 16:56
I tried assigning username = name in fn printUserName(), still don't have a global username variable – Programmer Dec 02 '17 at 17:06
does the alert shows the name? – Sajeetharan Dec 02 '17 at 17:07
within the function yes, but i need it out of the function. – Programmer Dec 02 '17 at 17:09
you need to understand the javascript basics first. I think you are from different background. you can use the variable within the scope and only if the variable value is assigned. Otherwise it will be undefined – Sajeetharan Dec 02 '17 at 17:10
yes i know, thats what i'm asking if there's any other workaround. putting it into another function is not helping. – Programmer Dec 02 '17 at 17:11

Javasctip variable not assigned

1 Answers1