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    copied!<p>The problem as @skparwal mentioned is that you are trying to insert '' into your primary key. You need to use <code>mysql_error()</code> after your insert to see what errors occur. Have a look at this link on how to do this: <a href="http://php.net/manual/en/function.mysql-error.php" rel="nofollow">http://php.net/manual/en/function.mysql-error.php</a></p> <p>Also, your ajax won't work as expected because you are not returning false for onClick() which means that the form is being submitted each time after the ajax call gets triggered.</p> <p>Furthermore, you should be moving to mysqli or PDO because mysql is deprecated.</p> <p><strong>Edit</strong></p> <p>Try this <strong>but it is not meant to work or be a final implementation,</strong> it is just to get you on the right path. It is untested so there might be some errors:</p> <p>Change your javascript to this:</p> <pre><code>$(function(){ $('#submit').click(postData); }); function postData() { var fName=$("#fName").val(); var lName=$("#lName").val(); var city=$("#city").val(); var data="fName="+fName+"&amp;lName="+lName+"&amp;city="+city+"&amp;submit=submit"; $.ajax({ type:'POST', url:'employeeajaxcodebehind.php', data:data, success:function(data){ $("#results").html(data); } }); return false; } </code></pre> <p>Change your PHP to this:</p> <pre><code>if(isset($_POST['submit'])) { $fName=$_POST['fName']; $lName=$_POST['lName']; $city=$_POST['city']; $Query="INSERT INTO employee(firstname,lastname,city) VALUES ('$fName','$lName','$city')"; if( !( $Result=mysql_query($Query) ) ) exit( '@todo need to check for error here' ); viewRecord(); } </code></pre> <p>Change your HTML to this:</p> <pre><code>&lt;form method="POST"&gt; Firstname: &lt;input type="text" id="fName" name="fName"&gt;&lt;br&gt; Lastname: &lt;input type="text" id="lName" name="lName"&gt;&lt;br&gt; City: &lt;input type="text" id="city" name="city"&gt;&lt;br&gt; &lt;input type="submit" id="submit" name="submit"&gt; &lt;/form&gt; </code></pre>
 

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