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Home/ Questions/Q 7588227
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T19:52:13+00:00 2026-05-30T19:52:13+00:00

I have: form.php preview.php form.php has a form in it with many dynamically created

  • 0

I have:
form.php
preview.php

form.php has a form in it with many dynamically created form objects. I use jquery.validation plugin to validate the form before submitting.
submit handler:

submitHandler: function() {
             var formData = $("#myForm").serialize();
            $.post("preview.php", {data: formData },function() {
    window.location.href = 'preview.php';
});

Question:
– How to change the current page to preview.php and show the data? my submitHandler doesnt work? Any tips?

preview.php:

$results = $_POST['data'];
        $perfs = explode("&", $results);
        foreach($perfs as $perf) {
            $perf_key_values = explode("=", $perf);
           $key = urldecode($perf_key_values[0]);
           $values = urldecode($perf_key_values[1]);
        }
        echo $key, $values;
    enter code here
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  1. Editorial Team
    Editorial Team
    2026-05-30T19:52:14+00:00Added an answer on May 30, 2026 at 7:52 pm

    I managed to solve my problem. without sessions.

    add to form:

     <form action="preview.php" onsubmit="return submitForPreview()">
     <input type="hidden" name="serial" id="serial" value="test">
    

    js:

    function submitForPreview()
    {
        if($("#form").valid()){
            $('#serial').val($("#newAdForm").serialize());
            return true;
        }else{
            return false;
        }
    
    }
    

    preview.php

    echo $_POST['serial']; 
    //Which shows the serialized string. YEEEEYYY :D 
    

    Thanks for help folk 😀

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