Why code for not run. JQUERY

Asked Jul 15 '15 at 09:16

Active Jul 15 '15 at 09:16

Viewed 36 times

I have a code

$('#ok').click(function(e) {
                    var number = $('#snumber').val();
            var scontent = $('#scontent').val();
            var img = $('#img').val();
            for(i = 0; i <= 2; i++) {
                dataEnd = 'du='+i+'&scontent='+scontent+'&img='+img;
                timeout = 9000*i + 9000;
                setTimeout(function(){
                    $.ajax({
                        type: 'POST',
                        url: 'xuly.php', 
                        data: dataEnd,
                        success: function(amw){
                            alert(dataEnd);
                        }
                    });
                }, timeout);
            }
            });

When I click id #ok. Code run but alert same du=2 every time. Plz tell me why???

asked Jul 15 '15 at 09:16

cuong ngo

2

`closures` is the answer – Tushar Jul 15 '15 at 09:16
I think my code is different so this code @T.J. Crowder – cuong ngo Jul 15 '15 at 09:57
No, it's the same thing. When the `success` function is called *later*, it sees the value of `dataEnd` as it is *then*, not when the function was created. So all the callbacks see the same `dataEnd` value (with `du=2`). – T.J. Crowder Jul 15 '15 at 10:12

Why code for not run. JQUERY

0 Answers0