jQuery ajax loop and iteration scope

Question

I\'m wondering why in the code below the i variable still shows \"5\" instead of showing \"1\" then \"2\" the

Answer 1

you can fix this using closures, wrapping the value of i:

for (var i = 0; i < 5; i++) {   
   (function(val){
      $.ajax({
          url: '/echo/html/',
          method:'post',
           data: {
               html: 'Ajax data'
           },
           success: function (resp) {
               $('#success').append(val);
           }
       })
       $('#outsideAjax').append(val); // is okay
     })(i);
}

Answer 2

Solution using Function#bind():
http://jsfiddle.net/RQncd/1/

for (var i = 0; i < 5; i++) {   
  $.ajax({
    url: '/echo/html/',
    method:'post',
    data: {
      html: 'Ajax data'
    },
    success: (function (i, resp) {
      $('#success').append(i);
    }).bind(null, i)
  });
  $('#outsideAjax').append(i);
}

Answer 3

This is due to Closures in JavaScript. Here's the fix -

for (var i = 0; i < 5; i++) { 
   (function(i){
    $.ajax({
        url: '/echo/html/',
        method:'post',
        data: {
            html: 'Ajax data'
        },
        success: function (resp) {
            $('#success').append(i) 
        }
    })
    })(i);

    $('#outsideAjax').append(i); 
}

Answer 4

fiddle Demo

var i = 0;

function ajax_call() {
    $.ajax({
        url: '/echo/html/',
        method: 'post',
        data: {
            html: 'Ajax data'
        },
        success: function (resp) {
            $('#success').append(i++);
            if (i < 5) {
                ajax_call();
            }
        }
    });
    $('#outsideAjax').append(i);
};
ajax_call();