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POload a php form in the same page using jquery and ajax
 

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    <p>I want a function that loads the values from a PHP form and show the result in a div in the same file ;</p> <p>I have a form in PHP file with the code bellow :</p> <pre><code>&lt;?php echo "&lt;script type='text/javascript' src='http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js'&gt;&lt;/script&gt;"; echo"&lt;script type='text/javascript' src='script.js'&gt;&lt;/script&gt;"; echo"&lt;form method='post' action='' &gt;"; echo " &lt;select name='mat'&gt;&lt;option value=''&gt;Choose: &lt;/option&gt;"; echo"&lt;option &gt;hello&lt;/option&gt;"; echo"&lt;/select&gt;"; echo"&lt;input type = 'submit' value='next' onclick='submit();' /&gt;"; echo"&lt;/form&gt;"; echo"&lt;div id ='result'&gt; &lt;/div&gt;"; ?&gt; </code></pre> <p>and the function in <code>script.js</code> is :</p> <pre><code>function submit() { $(function () { var mat = "mat"; var dataString = 'mat=' + mat; $.ajax({ type: "POST", url: "check.php", data: dataString, success: function (res) { goToByScroll("result"); $('#result').html("&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;").hide().fadeIn(2500, function () { $('#result').html(res + "&lt;br /&gt;&lt;br /&gt; Finished"); }); } }); return false; }); } </code></pre> <p>and the <code>check.php</code> that load the values is :</p> <pre><code>&lt;?php function safe($value) { htmlentities($value, ENT_QUOTES, 'utf-8'); return $value; } if (isset($_POST['mat'])) { $mat = safe($_POST['mat']); echo ($mat); } elseif (!isset($_POST['mat'])) { header('Location: error?mat=notfound'); } ?&gt; </code></pre> <p>But when I click next in the fist PHP file, it doesn't show anything, can you find the error?</p>
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