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Home/ Questions/Q 981301
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Editorial Team
  • 0
Editorial Team
Asked: May 16, 20262026-05-16T04:31:16+00:00 2026-05-16T04:31:16+00:00

I am using fancybox to open submitted form like this: <form name=form id=myForm action={$smarty.server.PHP_SELF}?action=abc

  • 0

I am using fancybox to open submitted form like this:

<form name="form" id="myForm" action="{$smarty.server.PHP_SELF}?action=abc" method="post" target="iframe" onsubmit="document.getElementById('iframe').src='{$smarty.server.PHP_SELF}?action=abc';">    
<input type="hidden" name="xyz" value="123">
<input type="submit" value="REGISTER" id="button" class="button">
</form> 

                $("#button").click(function() {                                            
                $("#iframe").fancybox({
                'titlePosition'     : 'inside',
                'transitionIn'      : 'fadeIn',
                'transitionOut'     : 'fadeOut'
                }).click();                            
            });

with <input type="submit" value="REGISTER" id="button" class="button"> everything is

working fine. but for some reason i need to use <a href="">REGISTER</a>. So I tried

<a href="#" onclick="document.getElementById('myForm').submit();">REGISTER</a>

&

<a href="javascript:document.form.submit();">REGISTER</a>

In above cases fancybox is popping out but my form is not getting submitted.

How do I submit my form successfully within click function?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T04:31:17+00:00Added an answer on May 16, 2026 at 4:31 am

    Create your link like this:

    <a href="#" id="submit_link">REGISTER</a>
    

    And use this code:

    $(function(){
      $('#submit_link').click(function(){
        $('#myForm').submit();
        return false;
      });
    });
    
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