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Editorial Team
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Asked: 2026-05-29T20:39:02+00:00

SO for some reason my form is submitting two times with a single button

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SO for some reason my form is submitting two times with a single button press. This is my first time using jquery Form plugin, and I imagine that jquery is submitting once and the form is “naturally” submitting as well. I have seen that the remedy is to attach a “return false” to the onSubmit event handler of the form. I thought I am doing that, but obviously it is not working.

Can anyone help?

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
<head>
<title>User form entry </title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script src="jquery.validate.js"></script>
<script src="jquery.passwordStrength.js"></script>
<script src="jquery.form.js"></script>
<script>

$(document).ready(function() {
    $('#signupForm').ajaxForm(function() { 
         var queryString = $('#signupForm').formSerialize(); 
         $.post('process.php', queryString);
    });
});
</script>
</head>
<body>
   <form id="signupForm" action="process.php" onsubmit="return false" method="post">
   <fieldset class="password">
          ... form goes here
   <button type="submit" name="formSubmit" value="submit">Click to submit</button>
   </form>

<div id="results"></div>

</body>
</html>

I have tried adding onsubmit=”return false” to the form element, but then I have no submission at all. I have also tried adding “return false;” to the jQuery, but then I still have double submissions. What am I missing? This seems to be the standard methodology according to the jQuery Form Plugin site.. how is my form different?

(By the way, just to be clear.. I am not talking about the problem of having multiple consecutive form submits by pressing the button repeatedly. My problem is “one submit button push = two submits”.)

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1 Answer

  1. Editorial Team
    $('#signupForm').ajaxForm(function() { 
            var queryString = $('#signupForm').formSerialize(); 
            $.post('process.php', queryString);

});

    This is submitting it twice …. the ajaxForm method once and then the post() the second time

    The ajaxForm method handles the form sumbission for you … you dont need to add the post() method … the function inside of ajaxForm is a callback, executed on success …

    $('#signupForm').ajaxForm(function() { 
       alert("Thank you for your comment!"); 
});

    this code would show the alert after the successful post … simple example here -> http://jquery.malsup.com/form/

    You should also remove the onsubmit attribute from the form …

    Update

    If you want to show results … do it like this :

    $(document).ready(function() { 
    $('#signupForm').ajaxForm({ target: '#results' }); 
    // this will output the responseText from the submitted form to the target DOM element
});
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