Question

可能重复：
Populating a dynamic drop down menu from a MySQL database

我的表单成功更新了mysql数据库，但是当用户重新登录时，我希望在下拉框中将数据库值指定为选定值，以便用户在之前选择的内容不起作用。

PHP：

<?php
session_start(); 
require_once("config.php"); 
if(!isset($_SESSION['username'])){
   header('Location: login.php'); /
    }
}
if(isset($_POST['submit'])){         
     $sql = "UPDATE user SET attendance1 = '" . mysql_real_escape_string($_POST['attendance1']) . "' WHERE username = '" .  mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());  
   $sql = "UPDATE user SET food1 = '" . mysql_real_escape_string($_POST['food1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());  
   $sql = "UPDATE user SET drink1 = '" . mysql_real_escape_string($_POST['drink1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());  

   header("Location: thanks.html", true, 303);
}
   $row2 = "SELECT * FROM user WHERE username = '" .       mysql_real_escape_string($_SESSION['username']) . "'";
     $result = mysql_query($row2) or die("Error in SQL: " . mysql_error());
     $row3 = mysql_fetch_array($result);
     echo $row3['shade1']; 
      ?>

FORM：

<form>

             <input name="attendance1" type="radio" id="Yes" value="Yes" checked="checked"/>Yes
             <br />
    <input name="attendance1" type="radio" id="No" value="No" />No
           </h3></td>
           <td>
                  <select name="colour1" id="colour1" >
               <option selected="selected">Please Select</option>
               <option>Red</option>
               <option>White</option>
               <option>Green</option>
             </select>
           </td>
           <td><h3>
                     <select name="shade1" id="shade1" >
               <option selected="selected">Please Select</option>
               <option value="Light" <?php if($row2['shade1']=="Light") { echo "selected"; }?>>Light</option>
               <option value="Heavy" <?php if($row2['shade1']=="Heavy") { echo "selected"; }?>>Heavy</option>
             </select>
           <td>&nbsp;</td>
           <td><label>
             <input type="submit" name="submit" id="button" value="Submit" />
           </label></td>
         </tr>
       </table>
</form>

Answer 1

您可以使用JQuery：

1）使用$.ajax中的$(document).ready()电话获取您想要的值

2）将它们设置为：

假设您的<select></select>元素的ID为“options”，请将其设置为当前选择的值，如下所示：

$("#options").val(optionToBeSelected);

其中optionToBeSelected是您要加入的value元素的<option></option>属性。

此代码应该适合您。如果有人在我的回答中看到主要/次要缺陷，请告诉我！

在下拉框中将数据库值指定为选定值

1 个答案: