Question

我是PHP的新手。我有mydropdown菜单程序的问题。我用名称id和名称创建表。与mysql数据库连接这就是我想要的， 当我在下拉菜单1中选择id 1然后下拉菜单2“仅显示”名称“ex：dk”和其他示例'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>

Answer 1

此

$result = mysql_query("SELECT * FROM cahkos Where name  = '.$row[id].' );

应该是这个

$result = mysql_query("SELECT * FROM cahkos Where name  = '".$row['id']."'" );

下拉菜单mysql PHP

1 个答案: