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[SOLVED] sql query
Posted: Sun Aug 08, 2004 10:30 pm
by g3ckO
Code: Select all
<?php
<?
include("database.php");
define('HOST', 'localhost');
define('USER', 'root');
define('PASS', '');
define('DB', 'db1');
$nama = $_POST['nama'];
$jawatan = $_POST['jawatan'];
$staffNo = $_POST['StaffNo'];
$unit = $_POST['unit'];
$jenis = $_POST['jenis'];
$datemula = $_POST['datemula'];
$datetamat = $_POST['datetamat'];
$jumlah = $_POST['jumlah'];
$datemohon = $_POST['datemohon'];
$RefID = "L(".$staffNo.")(".$datemohon.")";
global $conn;
$query = "INSERT INTO leave VALUES ('$nama', '$jawatan', '$staffNo', '$unit', '$jenis', '$datemula', '$datetamat', '$jumlah', '$jumlah',
'$datemohon', '$RefID')";
mysql_query($query,$conn);
?>
<br><br><center>YOUR APPLICATION HAVE BEEN SUBMITTED.</center>
?>
Why the data doesn't added to the database??
Posted: Sun Aug 08, 2004 11:12 pm
by infolock
this
Code: Select all
$query = "INSERT INTO leave VALUES ('$nama', '$jawatan', '$staffNo', '$unit', '$jenis', '$datemula', '$datetamat', '$jumlah', '$jumlah', '$datemohon', '$RefID')";
will not work because you are not telling mysql what fields you want the values to be inserted into..
ie, the query should look like this :
Code: Select all
$query = "INSERT INTO leave (field1, field2, field3, field4, field5, field6, field7, field8, field9, field10,field11) VALUES ('$nama', '$jawatan', '$staffNo', '$unit', '$jenis', '$datemula', '$datetamat', '$jumlah', '$jumlah', '$datemohon', '$RefID')";
Posted: Sun Aug 08, 2004 11:27 pm
by g3ckO
I do like this but still doesn't work:
Code: Select all
<?php
$query = "INSERT INTO leave (Nama, Jawatan, StaffNo, Unit, Jenis, DateMula, DateTamat, Jumlah, DateMohon, RefID) VALUES ('$nama',
'$jawatan', '$staffNo', '$unit', '$jenis', '$datemula', '$datetamat', '$jumlah', '$jumlah', '$datemohon', '$RefID')";
?>
Posted: Sun Aug 08, 2004 11:34 pm
by g3ckO
Sorry guys.. Its work now... Just some typing error..
Really sorry... TQ