因为下面的代码直接访问数据库表，但我希望它能在下拉菜单中显示表内容，就像我在一个下拉菜单中选择伊斯兰堡和在其他代码中给出的拉合尔一样，并按搜索按钮，然后它显示表航班。但是它直接显示
<p class="h2">Quick Search</p>
<div class="sb2_opts">
<p>
</p>
<form method="post" action="">
<p>Enter your source and destination.</p>
<p>
From:</p>
<select name="from">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<p>
To:</p>
<select name="To">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<input type="submit" value="search" />
</form>
</form> </table>
<?php
$from = isset($_POST['from'])?$_POST['from']:'';
$to = isset($_POST['to'])?$_POST['to']:'';
if( $from =='Islamabad'){
if($to == 'Lahore'){
$db_host = 'localhost';
$db_user = 'root';
$database = 'homedb';
$table = 'flights';
if (!mysql_connect($db_host, $db_user))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
die("Query to show fields from table failed");
}
$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h1>Table: {$table}</h1>";
echo "<table border='1'><tr>";
while($row = mysql_fetch_row($result))
{
echo "<tr>";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>'n";
}
}
}
mysqli_close($con);
?>
使用where语句。下面是创建where语句的通用方法
$where=array();
foreach(array('from'=>'from','To'=>'to') as $key=>$dbkey){
if(isset($_POST[$key]) && $_POST[$key]){
$where[]="`{$dbkey}`='".mysql_real_escape_string($_POST[$key])."'";
}
}
$result = mysql_query("SELECT * FROM {$table}".(count($where)?" WHERE (".implode(" && ",$where).")":""));