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Asked: 2026-05-24T05:20:52+00:00

I am completely new to javascript/jquery, and would appreciate any help. I am having

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I am completely new to javascript/jquery, and would appreciate any help.
I am having trouble with the function $.post because I am using radio in the form. I need to use the value of the chosen radio in a different file, so that I can process what should be outputted, and then I want to output something in place of where the form is.

Here is the form with type radio input:

<div id='poll'>   
<form name='poll_form' id='poll_form'>    
      <INPUT TYPE="radio" name='poll'  value ='poll1'/>Option1<br/>
      <INPUT TYPE="radio" name='poll'  value='poll2' />Option2<br/>
      <INPUT TYPE="radio" name='poll'  value='poll3'/>Option3<br/>
      <INPUT TYPE="radio" name='poll'  value='poll4'/>Option4</br>
      <INPUT TYPE='button'  value='Submit Vote' onClick="vote();" />
</form>
</div>

Here is the javascript/jquery to define the “vote();” function:

<head>
<script type = "text/javascript" src="jquery.js"></script>

<script type = "text/javascript">
  function vote() {
      $.post('file.php',$('input:radio[name=poll]:checked').val(),
        function(output){
          $("#poll").html(output).show();
        });
  };
</script>
</head>

Is $('input:radio[name=poll]:checked').val() the correct thing to use? And if so, how do I retrieve the value of $('input:radio[name=poll]:checked').val() in file.php?

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1 Answer

  1. Editorial Team

    To post values you would have to declare a post-variable and assign your value to it, i.e.

    $.post('file.php',{ poll: $('input:radio[name='poll']:checked').val() }, function() {
    $("#poll").html(output).show();
});

    In your PHP file you can access the value via

    $_POST['poll']
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