Convert $.post to $.ajax

Question

I want to convert my $.post call to $.ajax call because I need to clean the cache for every call.

My $.post call is as follows:

        $.post("test",   
            function(data) {
                $("#test").html(data);
                initTest();
            }
        ).success(function(){
            $('.box1').hide("slide", {direction: "left"}, 1000);
            $('.box3').show("slide", {direction: "right"}, 1000);
        });

I tried this but it doesn't work...

    $.ajax({
          type: "POST",
          url: "test",
          success: function (data) {
              $("#test").html(data);
              $('.box1').hide("slide", {direction: "left"}, 1000);
              $('.box3').show("slide", {direction: "right"}, 1000);
          },
          dataType: "json",
          cache: false
    });

Did you bother to read the [documentation](http://api.jquery.com/jQuery.post/)? It's the very first thing it mentions. — Jon, Apr 25 '12 at 11:20
what's the error? if you don't know look in your console or catch the error with an error function like this `error: function(xhr, ajaxOptions, thrownError){ alert(thrownError); }` — Fabian, Apr 25 '12 at 11:27

Denis Ermolin · Accepted Answer · 2012-04-25T11:29:35.820

$.ajax({
   method:"POST",
   url : "test.php",
   success : function(data) {
                $("#test").html(data);
                initTest();
                 $('.box1').hide("slide", {direction: "left"}, 1000);
     $('.box3').show("slide", {direction: "right"}, 1000);
            },
   cache : false
});

UPDATE I think you have problem in parsing data. In $.post by default is html dataType. In $.ajax call you changed it on "json". If no json is responsed so you have parse error and success handler dont call.

score 1 · Answer 2 · answered Apr 25 '12 at 11:22

 $.ajax('test')
    .done(function() {
      $('.box1').hide('slide', {direction: 'left'}, 1000);
      $('.box3').show('slide', {direction: 'right'}, 1000);
    })
    .fail(function() { alert('error'); })
    .always(function() { alert('complete'); });

Convert $.post to $.ajax

2 Answers2