Dropdown list data not match using php and mysql
Question
First of all thank you guys for helping me for this little problem I have.
Straight to the matters, but first of all I've been looking about this scripts by googling it and of course in this forum.
My Database:
(1). tbl_barang >
id_barang id_kategori id_klasifikasi nama_barang
(2). tbl_kategori >
id_kategori nama_kategori
(3). tbl_klasifikasi >
id_klasifikasi nama_klasifikasi
And here my scripts >
<!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>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<script language="javascript" type="text/javascript">
function showBarang(katid) {
document.frm.submit();
}
</script>
</head>
<body>
<form action="" method="post" name="frm" id="frm">
<table width="500" border="0">
<!--Kategori-->
<tr>
<td width="119">Kategori</td>
<td width="371">
<select name="id_kategori" id="id_kategori" onChange="showBarang(this.value);">
<option value="">--Select--</option>
<?php
$sql1="select * from tbl_kategori";
$sql_row1=mysql_query($sql1);
while($sql_res1=mysql_fetch_assoc($sql_row1))
{
?>
<option value="<?php echo $sql_res1["id_kategori"]; ?>" <?php if($sql_res1["id_kategori"]==$_REQUEST["id_kategori"]) { echo "Selected"; } ?>><?php echo $sql_res1["nama_kategori"]; ?></option>
<?php
}
?>
</select>
</td>
</tr>
<!-- Klasifikasi -->
<tr>
<td>Klasifikasi</td>
<td id="td_company">
<select name="id_klasifikasi" id="id_klasifikasi" onChange="showBarang(this.value);">
<option value="">--Select--</option>
<?php
$sql1="select * from tbl_klasifikasi";
$sql_row1=mysql_query($sql1);
while($sql_res1=mysql_fetch_assoc($sql_row1))
{
?>
<option value="<?php echo $sql_res1["id_klasifikasi"]; ?>" <?php if($sql_res1["id_klasifikasi"]==$_REQUEST["id_klasifikasi"]) { echo "Selected"; } ?>><?php echo $sql_res1["nama_klasifikasi"]; ?></option>
<?php
}
?>
</select>
</td>
</tr>
<tr>
<td>Nama Barang</td>
<td id="td_company">
<select name="id_barang" id="id_barang">
<option value="">--Select--</option>
<?php
$sql="select * from tbl_barang where id_kategori and id_klasifikasi='$_REQUEST[id_kategori] && [id_klasifikasi]'";
$sql_row=mysql_query($sql);
while($sql_res=mysql_fetch_assoc($sql_row))
{
?>
<option value="<?php echo $sql_res["id_barang"]; ?>"><?php echo $sql_res["nama_barang"]; ?></option>
<?php
}
?>
</select>
</td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
</table>
</form>
</body>
</html>
The problem is, when we choose kategori and choose klasifikasi the data 'nama_barang' should displays on the third dropdown list. But it's all mess, I don't know how to request two id's from the different tables and display it from one table 'tbl_barang'.
The short issue is, I don't know how to display the data from tbl_barang > 'nama_barang' that have related id's from 'tbl_kategori' and 'tbl_klasifikasi'.
Thank you so much if someone here help me, it's been three days and I'm stuck.
Thank you and best regards,
Kris
1 answers
solution1
0 ACCPTED 2014-01-22 10:07:38
I think you are trying to do a join between different tables.
Next point is your sql syntax:
$sql="select * from tbl_barang where id_kategori and id_klasifikasi='$_REQUEST[id_kategori] && [id_klasifikasi]'";
Try this one:
$sql="select * from tbl_barang where id_kategori = ".$_REQUEST[id_kategori]." and id_klasifikasi = ".$_REQUEST[id_klasifikasi].";";
When those ids are stored as numbers do not use quotation marks to wrap them.
Hope it helps. It is not really clear what is you problem.
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