Asynchronous AJAX call problem

Question

I have a jquery Ajax call function that I use to submit form to database to save. Also use that to get data from server and show to the user. When I post\save form data, I must show a result of the post to the user whether there was an error or not.

Everything works fine if I run in synchronous mode. I get the data if I want or get the result after posting using this single method.

But it seems not working in async mode.

How do I pass the data back to the caller? The web service returns the data correctly. See the alert() I have. It shows me the data.

How can I get data and show to user and post data and show result to user in async mode? Do I need to create 2 separate functions one for get and other to post?

Here is the function:

function MakeAjaxCall(svr, webmethod_name, op, btn, rslt, bGetData) {
   var data = "";
   var args = '';
   var l = arguments.length;
   if (l > 7) { for (var i = 7; i < l - 1; i += 2) { if (args.length != 0) args += ','; args += '"' + arguments[i] + '":"' + arguments[i + 1] + '"'; } }

   var surl = "http://<myserver>/" + svr + '.aspx';
   $.ajax({
      type: "POST",
      async: (op == "sync") ? false : true,
      url: surl + "/" + webmethod_name,
      data: "{" + args + "}",
      contentType: "application/json; charset=utf-8",
      dataType: "json",
      success: function ShowSuccess(res) {
         //alert('dfx res:'+res.d);//THIS SHOWS THE DATA FROM SQL CALL FINE
         data = res.d;
      },
      fail: function ShowFail(xhr) {         
         data = "ERROR";
      }
   });
}

Where is it failing? Have you looked at your console output to see if it's even reaching the database? What other debugging have you done? — JasCav
– JasCav, Commented Oct 4, 2010 at 17:55
@Proja How do you know the data isn't set correctly to your data variable after your success callback has run? Or is that not what you were trying to achieve? — bzlm
– bzlm, Commented Oct 4, 2010 at 18:16
data variable is getting set correctly. But the js function where I call MakeAjaxCall(), I need to have the data returned from web service (data variable gets it) — kheya
– kheya, Commented Oct 4, 2010 at 18:19
Basically the value of data variable needs to be passed to the caller of MakeAjaxCall() — kheya
– kheya, Commented Oct 4, 2010 at 18:20

jball · Accepted Answer · 2010-10-04 18:11:21Z

3

Add a callback function? E.g.,

function MakeAjaxCall(svr, webmethod_name, op, btn,
                      rslt, bGetData, callbackFunction) {
   //removed for brevity

   var surl = "http://<myserver>/" + svr + '.aspx';
   $.ajax({
      //removed for brevity
      success: function ShowSuccess(res) {
         //pass the data to the callbackFunction
         callbackFunction(res.d);
      },
      //etc.
   });

Then just pass the function name (or anonymous function) into the MakeAjaxCall function, like so:

//anonymous function
MakeAjaxCall(svrVal, somewebmethod_name, opVal, btnVal,
                      rsltVal, bGetDataVal, function(data) { alert(data); });

//named function
MakeAjaxCall(svrVal, somewebmethod_name, opVal, btnVal,
                      rsltVal, bGetDataVal, alert)

edited Oct 4, 2010 at 18:11

answered Oct 4, 2010 at 18:06

jball

25.1k9 gold badges72 silver badges92 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

kheya Over a year ago

How will this return the data to the caller js method?

jball Over a year ago

My point is that you should break up the caller method into a caller method and a callback method. The caller does everything up to the execution of MakeAjaxCall, and the callback method gets called on success, and passed the data it needs to finish everything up.

kheya Over a year ago

So you are asking to pass a function as a callback when I call MakeAjaxCall().

bzlm Over a year ago

@Proja If all else fails, try the async parameter in the ajax() call: api.jquery.com/jQuery.ajax

kheya Over a year ago

Do you have a sample code which shows the named function working? Where should the named function to be defined? On the caller page, right?

|

Collectives™ on Stack Overflow

Asynchronous AJAX call problem

1 Answer 1

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related