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Editorial Team
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Asked: 2026-06-07T20:56:18+00:00

I have this: <form method=post id=kl action=step2.php> <input type=radio name=rubrik value=bussines></input> <input type=radio namerubrik

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I have this:

<form method="post" id="kl" action="step2.php">

<input type="radio" name="rubrik" value="bussines"></input>
<input type="radio" name"rubrik" value="private"></input>

<input type="image" value="submit" src="/images/submit.png" alt="Submit" />

</form>

What i bassicaly want is: When the second radio button is checked, to submit the form to step2a.php, a different file. How can i do this? Jquery, Javascript, php?

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1 Answer

  1. Editorial Team

    You could do this with JavaScript (bind a submit listener that checks the value of the radio button and then sets the action property of the form), but it would be simpler and more reliable to do something (server side) along the lines of:

    <form ... action="step-selector.php">

    and

    <?php
    if (isset($_POST['rubrik']) && $_POST['rubrik'] == 'bussines') {
        include('step2.php');
    } elseif (isset($_POST['rubrik']) && $_POST['rubrik'] == 'private') {
         include('step2a.php');
    } else {
         include('error-state.php');
    }
 ?>
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