im new on PHP. i have problem on mydropdown menu program. i create table with name id and name .connected with mysql database this what i want, when i choose id 1 on dropdown menu 1 then dropdown menu 2 "show only" name "ex:dk" and other example 'id 2 " = name :"James " ....
<?php
/*---START -----*/
$host ='localhost';
$user ='root';
$pass ='';
$db ='sqldumb';
$kon =mysql_connect($host, $user, $pass);
if(!$kon)
die('EROR '.mysql_error());
$dbkon =mysql_select_db($db);
if(!$dbkon)
die('EROR '.mysql_error());
/*-------END ------*/
$sql = '';
$result = mysql_query("SELECT * FROM cahkos");
$sql .= '<select>';
while($row = mysql_fetch_array($result))
{
$sql .= "<option value='".$row['id']."'>".$row['id']."</option>";
}
$sql .= '</select>';
?>
<?php
$sql1 = '';
$result = mysql_query("SELECT * FROM cahkos Where name = '.$row[id].' );
$sql1 .= '<select>';
while($row = mysql_fetch_array($result))
{
$pilihan1 .= "<option value='".$row['name']."'>".$row['name']."</option>";
}
$sql1 .= '</select>';
?>
<html>
<head></head>
<body>
<table align=center >
<tr>
<td font-size=100>choose ID</td>
<td><?php echo $sql;?></td>
</tr>
<tr>
<td font-size=100>choose Name</td>
<td><?php echo $sql;?></td>
</tr>
</table>
</body>
This
$result = mysql_query("SELECT * FROM cahkos Where name = '.$row[id].' );
should be this
$result = mysql_query("SELECT * FROM cahkos Where name = '".$row['id']."'" );
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