xxreenaxx1 Posted February 16, 2011 Share Posted February 16, 2011 Hey guy <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("Examination", $con); $sql="INSERT INTO Test (Tes_Name, Tes_Description) VALUES ('$_POST[Tes_Name]','$_POST[Tes_Description]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> This is my code and I can enter Tes_Name on to mysql but nothing is shown for Tes_Description. The form is <html> <body> <h3>Test</h3> <table border="0"> <form method="POST" action="try2.php"> <tr><td>Tes_Name</td><td>:</td><td><input type="text" name="Tes_Name" size="20"></td></tr> <tr><td>Tes_Description</td><td>:</td><td><input type="Tes_Description" name="password" size="20"></td></tr> <tr><td> </td><td> </td><td><input type="submit" value="Submit"></td></tr> </form> </table> </body> </html> Quote Link to comment Share on other sites More sharing options...
Maq Posted February 16, 2011 Share Posted February 16, 2011 Because the name of your input field is "password" not "Tes_Description". Quote Link to comment Share on other sites More sharing options...
ansharma Posted February 16, 2011 Share Posted February 16, 2011 you supplied wrong argument in input field choose proper type field type="text" <input type="Tes_Description" name="password" size="20"> Quote Link to comment Share on other sites More sharing options...
xxreenaxx1 Posted February 16, 2011 Author Share Posted February 16, 2011 oops.. I though type was the name... Thank you Sorted Quote Link to comment Share on other sites More sharing options...
Maq Posted February 16, 2011 Share Posted February 16, 2011 In addition to what ansharma said, your input field should look like: Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.