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我需要一个从 mysql 加载数据并显示在下拉列表中的脚本。从那里，我需要将选定的数据传递到另一个页面。我已经完成了第一步。我现在可以从 mysql 表中加载数据并在下拉菜单中显示它们。具体如下。

<?php
include("config.php");
$result= mysql_query("SELECT folder_name FROM folders");

echo '<select name="directory">'; // Open your drop down box

while ($row = mysql_fetch_array($result)) {
   //echo "<option>" . $row['folder_name'] . "</option>";
      echo '<option value="'.$row['folder_name'].'">'.$row['folder_name'].'</option>';
}
echo '</select>';// Close your drop down box
?>

现在我需要帮助将选定的数据传递到另一个页面。请问有什么建议吗？

4

2 回答 2

1

让我考虑form发布到page2.phpfrompage1.php

page1.php

<form method="post" action="page2.php">
        //your dropdown code here
        <?php
              include("config.php");
              $result= mysql_query("SELECT folder_name FROM folders");

              $str = '';   
              $str .= '<select name="directory">'; // Open your drop down box

              while ($row = mysql_fetch_array($result)) {
              $str .= '<option    value="'.$row['folder_name'].'">'.$row['folder_name'].'</option>';
              }
              $str .= '</select>';// Close your drop down box

              echo $str;
         ?>

   <input type="submit" value="submit" />
</form>

在 page2.php 中，您可以访问dropdown所选值

$selVal = '';
if(isset($_POST['directory']))
{
    $selVal = $_POST['directory'];
}
于 2013-01-26T15:21:25.820 回答
1

创建一个 javascript 函数来处理带有文件夹名称数据的重定向： 

function changePage(folder){
 window.location.href= 'http://www.yourdomain.com/page2.php?folder=' + folder;
}

onchange 选项，以文件夹名称作为输入触发 changePage javascript 函数：

include("config.php");
$result= mysql_query("SELECT folder_name FROM folders");

echo '<select name="directory">'; // Open your drop down box

while ($row = mysql_fetch_array($result)) {
      echo '<option value="'.$row['folder_name'].'" onchange="changePage(\''.$row['folder_name'].'\')">'.$row['folder_name'].'</option>';
}
echo '</select>';// Close your drop down box

page2.php 

$folder_name = strip_tags($_GET['folder']);
于 2013-01-26T15:28:13.753 回答