Solved

ajax form submits twice

Posted on 2014-07-17
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1,065 Views
Last Modified: 2014-07-18
Experts, I've spent three frustrating days on a really simple procedure.  I need to evaluate a form that may change according to the selected option.  The class "required" points to the fields for validation.

My validator works great, but the form submits twice!

I've tried three solutions: event.preventDefault() and event.preventPropagation() ideas, return false -- nothing seems to work.  (See the attached js and php snippets.)

I've tried a number of plugins, but couldn't configure them to evaluate elements by their class.

Can you suggest anything else?  Any workarounds?  Do I need to try a different approach?

Pul-lease, help!
snippet.js
snippet.php
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Comment
Question by:Glen Gibb
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5 Comments
 
LVL 82

Expert Comment

by:leakim971
ID: 40203584
all the errors I saw about this is people still want to use a SUBMIT button (<input type="submit"...) instead a normal button (<input type="button"...)

I did not check your files but I'm sure you've one

The idea is to preventDefault and submit your form programatically instead using the submit button
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Author Comment

by:Glen Gibb
ID: 40203767
I'll give that a go.  Call submit on button click?
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LVL 13

Assisted Solution

by:duncanb7
duncanb7 earned 250 total points
ID: 40203924
Did you try to put e.preventDefault() and e.stopPropagation();  before $.ajax statement
and put $(this).unbind('submit') before "return false" (not "$(this).unbind('submit').submit();")
as the following code ?  Is it same ?

Duncan
e.preventDefault();
e.stopPropagation();
$.ajax (
			{	
				type: "post",
				url: $(this).attr('action'),
				data: $(this).serialize(),
				async: true,
				success: function(data)
				{
					//$(this).unbind('submit').submit();
					showResponse('success!');  
				},
				error: function(request, status, error, data)
				{
				alert('submit: ' + status + ', ' + error);
				} // end success ...		
			}); // end ajax ...
			$(this).unbind('submit');
                        return false;

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LVL 82

Accepted Solution

by:
leakim971 earned 250 total points
ID: 40204220
I'll give that a go.  Call submit on button click?

you need :

$("#fakeSubmitButton").click(function() {
  var form = $('#frm_client');
  if(validate('frm_client')) {
			$.ajax (
			{	
				type: "post",
				url: form.attr('action'),
				data: form.serialize(),
				async: true,
				success: function(data)
				{
                                     form.submit(); // this is the only place you submit it

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0
 

Author Closing Comment

by:Glen Gibb
ID: 40205236
Wow, I tried your suggestions with interesting results!

Leakim, I changed to a button.click and put a breakpoint on the form.submit.  When I ran the code, it hit the breakpoint only once.  But I got double the double - 4 sql commands for every item entered!

Duncan, I placed the prevents above the ajax, and the unbind just before the false, and now I'm back to just two sql commands for each item.

It seems that jQuery is not double-submitting; it must be the php.

I've checked this thoroughly but must have missed something.

I'd welcome any further suggestions, but thanks for your input.
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