I need to display data from a form in another form based on selection list/dropdown.
I keep getting an error, so I've included the code:
if($result === FALSE) {
die(mysql_error()); // Display a more meaningful error
}
while($row = mysql_fetch_array($result))
{
echo $row['NamaPelanggan'];
$row = mysql_fetch_array($sql);
$KodPelanggan = $row['KodPelanggan'];
$AlamatPelanggan = $row['AlamatPelanggan'];
$Bandar = $row['Bandar'];
$Negeri = $row['Negeri'];
$NoTelefon = $row['NoTelefon'];
$emel = $row['emel'];
}
But now, after I include that code, it shows Query was empty
. What does it mean? The original code is
if(isset($_POST['Submit'])) {
$sql = mysql_query ("INSERT INTO projek(idPel, nama_pegawai, jawatan_pegawai, notel_pegawai, hp_pegawai, nofaks_pegawai, emel_pegawai, kod_projek, norujukan, jenis_perolehan, tarikh_beli, harga_dokumen, kategori_projek, keterangan_projek, dok_proposal, status_projek, ada_invois, anggaran_kos, nilai_projek, kos_sebenar, margin_projek, info_plus, tarikh)
VALUES('$_POST[NamaPelanggan]','$_POST[nama_pegawai]','$_POST[jabatan_pegawai]','$_POST[notel_pegawai]','$_POST[hp_pegawai],'$_POST[nofaks_pegawai]','$_POST[emel_pegawai]','$_POST[kod_projek]','$_POST[norujukan]','$_POST[tarikh_beli]','$_POST[tarikh_tutup]','$_POST[harga_dokumen]','$_POST[kategori_projek]','$_POST[keterangan_projek]','$_POST[dok_proposal]','$_POST[status_projek]','$_POST[ada_invois]','$_POST[anggaran_kos]','$_POST[nilai_projek]','$_POST[kos_sebenar]','$_POST[margin_projek]','$_POST[info_plus]','$_POST[tarikh]')") or die(mysql_error());
header('Location: borang_projek.php');
}
else if(isset($_POST['NamaPelanggan'])) {
$NamaPelanggan=$_POST['NamaPelanggan'];
$sql = mysql_query("SELECT * FROM pelanggan WHERE id='$NamaPelanggan'");
$result = mysql_query($sql);
if($result === FALSE) {
die(mysql_error()); // Display a more meaningful error
}
while($row = mysql_fetch_array($result)) {
echo $row['NamaPelanggan'];
$row = mysql_fetch_array($sql);
$KodPelanggan = $row['KodPelanggan'];
$AlamatPelanggan = $row['AlamatPelanggan'];
$Bandar = $row['Bandar'];
$Negeri = $row['Negeri'];
$NoTelefon = $row['NoTelefon'];
$emel = $row['emel'];
}
}
else {
$NamaPelanggan="0";
$KodPelanggan = "";
$AlamatPelanggan = "";
$Bandar = "";
$Negeri = "";
$NoTelefon = "";
$emel = "";
}
How can I fix this?
I think your problem is in the else if where you call mysql_query twice.
$sql = mysql_query("SELECT * FROM pelanggan WHERE id='$NamaPelanggan'");
$result = mysql_query($sql);
The correct one should be:
$result = mysql_query("SELECT * FROM pelanggan WHERE id='$NamaPelanggan'");
i saw a lot of error in this code..
$sql = mysql_query ("INSERT INTO projek(idPel, nama_pegawai, jawatan_pegawai, notel_pegawai, hp_pegawai, nofaks_pegawai, emel_pegawai, kod_projek, norujukan, jenis_perolehan, tarikh_beli, harga_dokumen, kategori_projek, keterangan_projek, dok_proposal, status_projek, ada_invois, anggaran_kos, nilai_projek, kos_sebenar, margin_projek, info_plus, tarikh)
VALUES('$_POST[NamaPelanggan]','$_POST[nama_pegawai]','$_POST[jabatan_pegawai]','$_POST[notel_pegawai]','$_POST[hp_pegawai],'$_POST[nofaks_pegawai]','$_POST[emel_pegawai]','$_POST[kod_projek]','$_POST[norujukan]','$_POST[tarikh_beli]','$_POST[tarikh_tutup]','$_POST[harga_dokumen]','$_POST[kategori_projek]','$_POST[keterangan_projek]','$_POST[dok_proposal]','$_POST[status_projek]','$_POST[ada_invois]','$_POST[anggaran_kos]','$_POST[nilai_projek]','$_POST[kos_sebenar]','$_POST[margin_projek]','$_POST[info_plus]','$_POST[tarikh]')") or die(mysql_error());
it should be
$sql = mysql_query ("INSERT INTO `projek`(`idPel`, `nama_pegawai`, `jawatan_pegawai`, `notel_pegawai`, `hp_pegawai`, `nofaks_pegawai`, `emel_pegawai`, `kod_projek`, `norujukan`, `jenis_perolehan`, `tarikh_beli`, `harga_dokumen`, `kategori_projek`, `keterangan_projek`, `dok_proposal`, `status_projek`, `ada_invois`, `anggaran_kos`, `nilai_projek`, `kos_sebenar`, `margin_projek`, `info_plus`, `tarikh`)
in your values, the post name should inside the quote, you have incorrect systax, do this sample code
VALUES('".$_POST['NamaPelanggan']."','".$_POST['nama_pegawai']."'etc);
//the name of $_POST should inside of quote..do you realize now what your error?
and in your else it should be
$sql = "SELECT * FROM pelanggan WHERE id='$NamaPelanggan'";
$result = mysql_query($sql);