php代码没有使用wamp服务器从mysql数据库中获取数据

Question

我想在 phpMyAdmin 中显示来自数据库的表，方法是在下拉菜单中的每个不同选项中通过按下搜索按钮显示来自数据库的不同表。但它没有这样做。

<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
$con=mysqli_connect("localhost","root","","test"); 
if (mysqli_connect_errno()) 
{ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
}

if(isset($_POST['from']) and isset($_POST['To'])) {
    $from = $_POST['from'] ;
    $to = $_POST['To'] ;
    $table = array($from, $to);

    switch ($table) {
        case array ("Islamabad", "Lahore") :
           $result = mysqli_query($con,"SELECT * FROM flights");
           echo "</flights>";      //table name is flights
           break;
        case array ("Islamabad", "Murree") :
           $result = mysqli_query($con,"SELECT * FROM isb to murree");
          echo "</isb to murree>";     //table name isb to murree
          ;
          break;
       case array ("Islamabad", "Muzaffarabad") :
           $result = mysqli_query($con,"SELECT * FROM isb to muzz");
           echo "</isb to muzz>";
           break;
       //.....
       //......
       default:
          echo "Your choice is nor valid !!";
    }

}
mysqli_close($con);
?>

Answer 1

您当前遇到的问题似乎是由您的表名引起的。当它们包含空格（或者是保留字、以数字开头等）时，需要用反引号引起来：

SELECT * FROM `isb to murree`

但是，在我继续之前，我可能会好好看看数据库结构（和表名......），因为每个目标组合都有一个不同的表并不能很好地扩展。您的所有组合都可以轻松存储在一个表中（仅使用数字来识别机场），可能还需要一些额外的列来获取更多信息。