帮助jQuery Ajax成功事件

Question

I'm having an issue with my update button and jquery ajax. Right now when I click on my update button, it saves whatever updated data to the database. My goal is I want to slide up a message if the update is successful. I was looking at ajax post and using the success event seems like it would work but I dont know how to incorporte it. How would I do this? Would it be something like this?

     $(document).ready(function(){
        $('#divSuccess').hide();

        $('#btnUpdate').click( function() {
        alert('button click');
            $.ajax({ 
                  url: "test.aspx", 
                  context: document.body, 
                  success: function(){ 
                    $('#divSuccess').show("slide", { direction: "down" }, 3000);
                    $('#divSuccess').hide("slide", { direction: "down"}, 5000);
                  }
                }); 
        });
    });

Answer 1

check out this question for an example on how to handle the success event. Hope this helps!

Answer 2

$("#targetDiv").load("page.php",$("#form").serializeArray(),function (response) 
            {
              if (response == '0' && response != '')
               alert('Request not sent to server !\n');
              else if(response == '-1')
               alert('Please write some more !\n');
              else
              {
                alert("success! ");
              }
            }
         );

i've echo ed 0 and -1 for failure and other for success

Answer 3

In the jquery post function, you can execute some callback function.

    function (data, textStatus) {
      // data could be xmlDoc, jsonObj, html, text, etc...
      this; // the options for this ajax request
      // textStatus can be one of:
      //   "timeout"
      //   "error"
      //   "notmodified"
      //   "success"
      //   "parsererror" 
      // NOTE: Apparently, only "success" is returned when you make
      // an Ajax call in this way. Other errors silently fail.
      // See above note about using $.ajax.
    }

http://docs.jquery.com/Post

Answer 4

With at least jQuery 1.5, you've got deferred objects and new syntax for AJAX events (including success ).

var $ajaxcall = $.ajax({
    url : 'myurl.svc/somemethod',
    data : '{ somedata : "sometext" }'
});

$ajaxcall.success(function() {
    // do something on successful AJAX completion
});

Of course you can chain that as well, and call something along the lines of $.ajax().success() or something.

Just wrote a blog post on it myself , if you're interested in reading more.