Use variable out of ajax

Question

How I can use count variable out of ajax? In ajax function "count" shows count, but dessous nothing.

var count;
$.ajax({
cache   :   false,
dataType:   'json',
type    :   "POST",
url     :   "count.php",
success  :   function(tdata){
count = tdata;
console.log(count); //this works

}               

});

console.log(count); //this doesn't work

what do you mean it doesn't work ? Does it say "undefined" ? — Bdloul, Jul 19 '17 at 05:32
Your ajax call is asynchronous, so your console log getting print before response, you can make it sync but it is not good practice. — Nitin Dhomse, Jul 19 '17 at 05:33
Possible duplicate of [How do I return the response from an asynchronous call?](https://stackoverflow.com/questions/14220321/how-do-i-return-the-response-from-an-asynchronous-call) — t.niese, Jul 19 '17 at 05:34
Or [Why is my variable unaltered after I modify it inside of a function asynchrony](https://stackoverflow.com/questions/23667086/) — t.niese, Jul 19 '17 at 05:38

score 1 · Answer 1 · answered Jul 19 '17 at 05:35

1

$.ajax() is async, you need to wait for it to finish.

var count;
$.ajax({
cache   :   false,
dataType:   'json',
type    :   "POST",
url     :   "count.php",
success  :   function(tdata){
    count = tdata;
    console.log(count); //this works

}               

})
.done(() => {
    // this code runs after ajax is resolved
    console.log(count);
});

Refer to http://api.jquery.com/jQuery.ajax/ for other chaining methods

answered Jul 19 '17 at 05:35

Sean

337
2
7

But I have big code! I can;t put it .done(() => {}); – PHP Useless Jul 19 '17 at 05:41
Size of the code doesn't matter. Also, depending on your environment, you could try using async/await. – Sean Jul 19 '17 at 05:44

Use variable out of ajax

1 Answers1