maintain function order in jQuery ajax callback function

Question

I have a jQuery ajax function. the callback function consists of two functions:

1) Take the ajax result (which is an html form) and insert it as an html span's inner html. 2) Submit this form

How can I ensure that the form is loaded before JavaScript tries to submit it?

Code:

$("select").live("change", function(){
  FormUpdate();
})

function FormUpdate() {

  $.ajax({
  type: "POST",
  url: "index.php?UpdateForm=Yes",
  data: $("#Form").serialize(),
  success: function(msg){
    $("span#Content").html(msg);
    $("#Form").submit();
  }
  });

}

My problem comes because javascript tries to submit the form before it has loaded and my data submitted.

mauris · Accepted Answer · 2010-01-07 23:17:09Z

1

Just simply put the function for taking ajax result and insert into the DOM in front and the form submission function at the back.

Only after insertion is done, the next function will be called.

Example:

$.ajax({
   type: "POST",
   url: "backend.php",
   data: "name=John",
   dataType: "json"
   success: function(msg){
     $("#thediv").html(msg.html);
     $("form#theform").submit();
   }
 });

answered Jan 7, 2010 at 23:17

mauris

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1 Comment

Brian Over a year ago

I guess I have another issue with my code if the callback function will be synchronous

rfunduk · Accepted Answer · 2010-01-07 23:04:03Z

0

DOM manipulations are synchronous so the form wont be submitted until afterwards. Are you having a specific problem with this scenario? Please post tons of code, if so :)

answered Jan 7, 2010 at 23:04

rfunduk

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1 Comment

Nakul Over a year ago

right, if you can't see it happening (it would be lightening fast :P), you can try putting $('#Form').submit() call in a timeout() function

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