如何从下拉列表中回显sql语句的结果

Question

您好我想回复菜单价格的结果。下面是代码，但在选择菜单类型（如菜单1）后，没有显示任何内容。

 <select name="updateMenuPrice" class="StyleTextField" id="updateMenuPrice">
                <option value="-1">Select One</option>
                <option value="Menu 1">Menu 1</option>
                <option value="Menu 2">Menu 2</option>
                <option value="Menu 3">Menu 3</option>
              </select>

 <?php
          if(isset($_POST['UpdatePriceBtn'])){

           $_SESSION["cupdateprice"] = $_POST['updateMenuPrice']; 


       }


 $con = mysqli_connect("localhost", "root", "password", "test") or die(mysql_error());
            if (!$con) {
            die('Could not connect: ' . mysql_error());
          echo 'fail1\n';
          }//if
       // else echo 'Connected successfully';

          $sql =  "SELECT Price from menu where Menu_type = '".$_SESSION["cupdateprice"]."'";

           if(mysqli_query($con, $sql)){
                 $result = mysqli_query($con, $sql);
                echo  $sql ;
                       } 
                    else {
                      echo "failed";

                    }
          ?>