pencarian data 3 tabel berlelasi dalam mysql
maba · Nov 16, 2012
tb.user
id_user
usermane
password
tabel 2 alumni
id_alumni
id_user
id_tempat_kerja
nama_lengkap
alamat
tlp
tahun_lulus
tabel 3 tempat_kerja
id_tempat
nama_instansi
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<link rel="stylesheet" href="Css/style.css" type="text/css">
<link rel="stylesheet" href="Css/form.css" type="text/css">
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>PENCARIAN SEDERHANA</title>
</head>
<body>
<div><?php require_once ("../../fungsi/data.php") ?></div>
<hr class="garis"><br />
<?php require_once ("../../koneksi/koneksi.php") ?>
<table width="125%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#11A2E8">
<tr>
<td bgcolor="#D6EFFC"><br /><br />
<div style="font-size:14px; text-align:center">PENCARIAN ALUMNI BERDASARKAN KATEGORI :</div>
<div align="center"><br>
<li type="circle">Nama Alumni</li><br>
<li type="circle">Tahun Lulus</li>
</div>
<br /><form id="form1" name="form1" method="post" action="">
<div align="center">
<input name="cari" type="text" id="cari" size="50" />
<input type="submit" name="Submit" value="Cari" class="submit" />
</div>
</form> <br /><br />
<?
$cari=$_POST;
if(!empty($cari)){
?>
<p align="center"><strong>HASIL PENCARIAN</strong> :</p>
<br />
<p><?
$dataalumni=mysql_query("SELECT t_a.id_alumni, t_a.nama_lengkap, t_a.alamat, t_a.no_tlp, t_a.tahun_lulus, t_a.foto, t_k.nama_instansi, tu.username
FROM alumni t_a
INNER JOIN tempat_kerja t_k ON t_k.id_tempat_kerja = t_a.id_tempat_kerja INNER JOIN t_user tu ON tu.id_user = tu.id_user WHERE tahun_lulusan='$cari' or nama_lengkap LIKE '%$cari%'");
$totalseluruhalumni=mysql_query($dataalumni);
if($totalseluruhalumni=='0'){
echo "<center><blink><font color='red'>Maaf Data yang anda cari tidak ada di dalam database</font></bink></center><br />";
} else {
echo "<br /><div style='border:0px solid #CCC; padding:5px;'>Jumlah Seluruh Data yang ditemukan Adalah <b>$totalseluruhpendaftar</b> Data</div>";
}
?>
<table width="98%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#D78EFA">
<tr>
<td height="36" align="center" valign="middle" bgcolor="#F0D7FD"><strong>No</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Nama</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Jenis Kelamin </strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Npm</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Alamat</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Lihat Data Lengkap</strong></td>
</tr>
<?
while($dataalumni1=mysql_fetch_object($dataalumni)){
?>
<tr>
<td height="25" bgcolor="#FFFFFF"><?=$dataalumni1[no]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[nama_lengkap]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[alamat]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[no_tlp]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[tahun_lulus]?></td>
<td bgcolor="#FFFFFF"><a href='lihatdatadetail.php?nama=<?=$dataalumni1[nama]?>'>Detail</a></td>
<?
}
?>
</table>
</p><br /><br />
<?
}
?>
</td>
</tr>
</table>
</body>
</html
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\alumni\modul\mod_alumni\listing.php on line 55
mohon bantuanya mas...apa yang kurang ya. dalam script ini. kutunggu jawabanya...
id_user
usermane
password
tabel 2 alumni
id_alumni
id_user
id_tempat_kerja
nama_lengkap
alamat
tlp
tahun_lulus
tabel 3 tempat_kerja
id_tempat
nama_instansi
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<link rel="stylesheet" href="Css/style.css" type="text/css">
<link rel="stylesheet" href="Css/form.css" type="text/css">
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>PENCARIAN SEDERHANA</title>
</head>
<body>
<div><?php require_once ("../../fungsi/data.php") ?></div>
<hr class="garis"><br />
<?php require_once ("../../koneksi/koneksi.php") ?>
<table width="125%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#11A2E8">
<tr>
<td bgcolor="#D6EFFC"><br /><br />
<div style="font-size:14px; text-align:center">PENCARIAN ALUMNI BERDASARKAN KATEGORI :</div>
<div align="center"><br>
<li type="circle">Nama Alumni</li><br>
<li type="circle">Tahun Lulus</li>
</div>
<br /><form id="form1" name="form1" method="post" action="">
<div align="center">
<input name="cari" type="text" id="cari" size="50" />
<input type="submit" name="Submit" value="Cari" class="submit" />
</div>
</form> <br /><br />
<?
$cari=$_POST;
if(!empty($cari)){
?>
<p align="center"><strong>HASIL PENCARIAN</strong> :</p>
<br />
<p><?
$dataalumni=mysql_query("SELECT t_a.id_alumni, t_a.nama_lengkap, t_a.alamat, t_a.no_tlp, t_a.tahun_lulus, t_a.foto, t_k.nama_instansi, tu.username
FROM alumni t_a
INNER JOIN tempat_kerja t_k ON t_k.id_tempat_kerja = t_a.id_tempat_kerja INNER JOIN t_user tu ON tu.id_user = tu.id_user WHERE tahun_lulusan='$cari' or nama_lengkap LIKE '%$cari%'");
$totalseluruhalumni=mysql_query($dataalumni);
if($totalseluruhalumni=='0'){
echo "<center><blink><font color='red'>Maaf Data yang anda cari tidak ada di dalam database</font></bink></center><br />";
} else {
echo "<br /><div style='border:0px solid #CCC; padding:5px;'>Jumlah Seluruh Data yang ditemukan Adalah <b>$totalseluruhpendaftar</b> Data</div>";
}
?>
<table width="98%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#D78EFA">
<tr>
<td height="36" align="center" valign="middle" bgcolor="#F0D7FD"><strong>No</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Nama</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Jenis Kelamin </strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Npm</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Alamat</strong></td>
<td align="center" valign="middle" bgcolor="#F0D7FD"><strong>Lihat Data Lengkap</strong></td>
</tr>
<?
while($dataalumni1=mysql_fetch_object($dataalumni)){
?>
<tr>
<td height="25" bgcolor="#FFFFFF"><?=$dataalumni1[no]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[nama_lengkap]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[alamat]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[no_tlp]?></td>
<td bgcolor="#FFFFFF"><?=$dataalumni1[tahun_lulus]?></td>
<td bgcolor="#FFFFFF"><a href='lihatdatadetail.php?nama=<?=$dataalumni1[nama]?>'>Detail</a></td>
<?
}
?>
</table>
</p><br /><br />
<?
}
?>
</td>
</tr>
</table>
</body>
</html
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\alumni\modul\mod_alumni\listing.php on line 55
mohon bantuanya mas...apa yang kurang ya. dalam script ini. kutunggu jawabanya...
Silahkan login untuk menjawab!
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Ellyx Christian
· Nov 19, 2012
· 0 Suka
· 0 Tidak Suka
kalau errornya seperti "supplied argument is not a valid MySQL result resource" pasti errornya karena salah sql query. Coba:
$dataalumni=mysql_query("SELECT t_a.id_alumni, t_a.nama_lengkap, t_a.alamat, t_a.no_tlp, t_a.tahun_lulus, t_a.foto, t_k.nama_instansi, tu.username
FROM alumni t_a
INNER JOIN tempat_kerja t_k ON t_k.id_tempat = t_a.id_tempat_kerja INNER JOIN t_user tu ON tu.id_user = tu.id_user WHERE tahun_lulusan='$cari' or nama_lengkap LIKE '%$cari%'");
$dataalumni=mysql_query("SELECT t_a.id_alumni, t_a.nama_lengkap, t_a.alamat, t_a.no_tlp, t_a.tahun_lulus, t_a.foto, t_k.nama_instansi, tu.username
FROM alumni t_a
INNER JOIN tempat_kerja t_k ON t_k.id_tempat = t_a.id_tempat_kerja INNER JOIN t_user tu ON tu.id_user = tu.id_user WHERE tahun_lulusan='$cari' or nama_lengkap LIKE '%$cari%'");