[ASK] Display data from selected dropdown list

cubluwt · January 21, 2014

Hi, I'm new here. But many posts here that helped me out before. Thanks alot to this forum, but now I need some help from you guys.

First of all this is my database, you can copy and paste it to notepad and save as .sql file to import to mysql.

-- phpMyAdmin SQL Dump

-- version 3.2.4

-- http://www.phpmyadmin.net

--

-- Host: localhost

-- Generation Time: Jan 21, 2014 at 01:14 AM

-- Server version: 5.1.41

-- PHP Version: 5.3.1



SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";





/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;

/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;

/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;

/*!40101 SET NAMES utf8 */;



--

-- Database: `db_cobadulu`

--



-- --------------------------------------------------------



--

-- Table structure for table `tbl_barang`

--



CREATE TABLE IF NOT EXISTS `tbl_barang` (

  `id_barang` int(11) NOT NULL AUTO_INCREMENT,

  `id_kategori` int(11) NOT NULL,

  `id_kalsifikasi` int(11) NOT NULL,

  `nama_barang` varchar(128) NOT NULL,

  PRIMARY KEY (`id_barang`)

) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;



--

-- Dumping data for table `tbl_barang`

--



INSERT INTO `tbl_barang` (`id_barang`, `id_kategori`, `id_kalsifikasi`, `nama_barang`) VALUES

(1, 2, 3, 'Samsung S4'),

(2, 3, 2, 'Vas Bunga'),

(3, 4, 1, 'Mug Porselen');



-- --------------------------------------------------------



--

-- Table structure for table `tbl_kategori`

--



CREATE TABLE IF NOT EXISTS `tbl_kategori` (

  `id_kategori` int(11) NOT NULL AUTO_INCREMENT,

  `nama_kategori` varchar(128) NOT NULL,

  PRIMARY KEY (`id_kategori`)

) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;



--

-- Dumping data for table `tbl_kategori`

--



INSERT INTO `tbl_kategori` (`id_kategori`, `nama_kategori`) VALUES

(1, 'Elektronik'),

(2, 'Pecah Belah'),

(3, 'Gift'),

(4, 'Kerajinan');



-- --------------------------------------------------------



--

-- Table structure for table `tbl_klasifikasi`

--



CREATE TABLE IF NOT EXISTS `tbl_klasifikasi` (

  `id_klasifikasi` int(11) NOT NULL AUTO_INCREMENT,

  `nama_klasifikasi` varchar(128) NOT NULL,

  PRIMARY KEY (`id_klasifikasi`)

) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;



--

-- Dumping data for table `tbl_klasifikasi`

--



INSERT INTO `tbl_klasifikasi` (`id_klasifikasi`, `nama_klasifikasi`) VALUES

(1, 'Ringan'),

(2, 'Berat'),

(3, 'Cukup Berat'),

(4, 'Sangat Berat');



/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;

/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;

/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;

And then this is the script I have:

<script languange="Javascript1.2">
function pilih(id){
	location.replace("index.php?id="+id);
}
</script>

<?php
$host="localhost";
$user="root";
$pass="";
$db="db_cobadulu";

//koneksi 
$koneksi=mysql_connect($host,$user,$pass);
mysql_select_db($db,$koneksi);

if($_GET['']!==""){
	$id=$_GET['id'];
	
	
	$query=mysql_query("SELECT b.*, nama_barang, p.nama_kategori, kl.nama_klasifikasi FROM tbl_barang AS b
						LEFT JOIN tbl_kategori AS p ON b.id_kategori=p.id_kategori
						LEFT JOIN tbl_klasifikasi AS kl ON b.id_klasifikasi=kl.id_klasifikasi
						WHERE id_barang='$id'");
	
	?>
	<table border="1">
	<tr><th>No</th><th>Nama Barang</th><th>Kategori</th><th>Klasifikasi</th>
	<?php
	//while($row=mysql_fetch_array($query) or die (mysql_error())){
	while($row=mysql_fetch_array($query)){
		?>
		<tr><td><?php echo $c=$c+1;?></td><td><?php echo $row['nama_barang'];?></td><td><?php echo $row['nama_kategori']; ?></td><td><?php echo $row['nama_klasifikasi']; ?></td>
		</tr>
		<?php
	}
	?></table><?php
}

?>
<html>
<head><title>Gudang Barang</title></head>
<body>
<p> </p>
<table border="0">
<tr>
</tr>
<tr>
	<td width="195" valign="top">
	<select name="id" id="id" onChange="pilih(this.value)">
		<option value="0" selected="selected">Pilih Kategori</option>
		<?php 
		$query_limit=mysql_query("select * from tbl_kategori");
		
		while($row=mysql_fetch_array($query_limit))
		{
			?><option value="<?php  echo $row['id_kategori']; ?>"><?php  echo $row['nama_kategori']; ?></option><?php 
		}
		?>
	</select>	
	</td>
	<td width="195" valign="top">
	<select name="id" id="id" onChange="pilih(this.value)">
		<option value="0" selected="selected">Pilih Klasifikasi</option>
		<?php 
		$query_limit=mysql_query("select * from tbl_klasifikasi");
		
		while($row=mysql_fetch_array($query_limit))
		{
			?><option value="<?php  echo $row['id_klasifikasi']; ?>"><?php  echo $row['nama_klasifikasi']; ?></option><?php 
		}
		?>
	</select>	
	</td>
</tr>
</table>
</body>
</html>

The problem is, the output is misserable. Yes it did showing the data from two tables. But it's a mess. What I'm saying is, I intend to show/display the data from table "tbl_barang" that connected with two tables "tbl_kategori" and "tbl_klasifikasi" by choosing data from the dropdown list. But not all the data came out, and also when I was putting the "submit" button it doesn't really work at all.

I was looking the old posts here, I've tried googling it but no results. Most of the results is "How to displaying data from database to dropdown list", but what I needed is "How to display data from selected item/data from the dropdown list". I need ur help guys...I cannot sleep, it's always in my mind...

Thank you so much for the helps.

Best regards,

Kris

Barand · January 21, 2014

First you only want to run the search query when

if(isset($_GET['id']) && $_GET['id']!==""){...

Second, your search query fails because of a spelling error in the table definition

CREATE TABLE IF NOT EXISTS `tbl_barang` (

  `id_barang` int(11) NOT NULL AUTO_INCREMENT,

  `id_kategori` int(11) NOT NULL,

  `id_kalsifikasi` int(11) NOT NULL,                                <---- should be klasifikasi

  `nama_barang` varchar(128) NOT NULL,

  PRIMARY KEY (`id_barang`)

)

Third and more serious

You have two dropdowns

one gives the classification id with name = "id"

other gives category id with name = "id"

So how do you know if category or classification was selected? they need to pass separate names in your $_GET.

However, no matter which id is selected you search for "WHERE id_barang='$id' ". You need to search for whichever id was passed in the $_GET (category or classification)

cubluwt · January 21, 2014

First you only want to run the search query when
if(isset($_GET['id']) && $_GET['id']!==""){...
Second, your search query fails because of a spelling error in the table definition
CREATE TABLE IF NOT EXISTS `tbl_barang` (

  `id_barang` int(11) NOT NULL AUTO_INCREMENT,

  `id_kategori` int(11) NOT NULL,

  `id_kalsifikasi` int(11) NOT NULL,                                <---- should be klasifikasi

  `nama_barang` varchar(128) NOT NULL,

  PRIMARY KEY (`id_barang`)

)
Third and more serious

You have two dropdowns

one gives the classification id with name = "id"

other gives category id with name = "id"

So how do you know if category or classification was selected? they need to pass separate names in your $_GET.

However, no matter which id is selected you search for "WHERE id_barang='$id' ". You need to search for whichever id was passed in the $_GET (category or classification)

Thank you so much...I get it...I get it... :happy-04:

Let me try it first...

About the id_kalsifikasi yes, I just found it after I wrote this posts and already fix it.... :tease-03:

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[ASK] Display data from selected dropdown list

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