please help me how to find the solution for the error above...below is showing my coding...

<?php
if (isset($_REQUEST["submit"]))
{
$Id=$_REQUEST['Id'];
$NAMA=$_REQUEST['NAMA'];
$NOMBOR_ID=$_REQUEST['NOMBOR_ID'];
$JURUSAN=$_REQUEST['JURUSAN'];

$dbhost="localhost";
$dbuser="root";
$dbpass="";
$db_name="jmti";

$connect = mysql_connect($dbhost,$dbuser,$dbpass);
$selectdb = mysql_select_db($db,$connect);
$checkid = mysql_query("select * from daftar where NAMA = '$NAMA'");

if(mysql_num_rows($checkid)==0)
{$save = "insert into daftar(Id,NAMA,NOMBOR_ID,JURUSAN) values
('$Id','$NAMA','$NOMBOR_ID','$JURUSAN')";

mysql_query($save) or die ("query error!");
echo "<p>Thank you for using our service, <b>$NAME</b>,your are choosed JURUSAN <b>$JURUSAN</b>";
}
else
echo"<br>Duplicate Food ID";
}
?>

please help me..

Do you get the same error if you query select * from daftar where NAMA = '$NAMA'" is launched via phpmyadmin directly?

Do you really use correct table/field names?

YUP...i use the correct table n field...My table's name is daftar and their fields are NAMA and JURUSAN...

Try:

PHP Code:

$connect = mysql_connect($dbhost,$dbuser,$dbpass);
if (!$connect) {die('Failed to connect to Server! - '.mysql_error());}

$selectdb = mysql_select_db($db,$connect);
if (!$selectdb) {die('Failed to connect to DB! - '.mysql_error());}

    $checkid = mysql_query("select * from daftar where NAMA = '$NAMA'");
if (!$checkid) {die('Failed to execute query! - '.mysql_error());} 


This way you can better troubleshoot.

T.Q...i will try.....

Any good news?