我有两个表页面和上传
页码：
page_id, page_name, page_parent
( 4, test, 0)
( 5, test-sub, 4)
( 6, test-sub2, 4)
上传：
upload_id, upload_name, upload_root
(1, test, 4)
(2, test-sub, 4)
(3, test-sub2, 4)
我很成功地将page_id输入到upload_root中，但当页面有子菜单并输入到page_parent时遇到了问题。
我的插入命令如下所示：
$name = $_POST['name'];
$root = $_POST['root'];
$insert_sql = "INSERT INTO tbl_upload VALUES ('' , '"$name'", '"$root'")";
$root = $_POST['page_id'];?>
<form action="" enctype="multipart/form-data" method="post">
<div>Name: <input name="title" type="text" size="50"/></div>
<input name="page_id" type="hidden" value= "<?php echo"$page_id"; ?>"
<input type="submit" value="Upload" id="upload" class="upload"/>
</form>
我的fetch命令如下所示：
$fetch_sql = "SELECT upload_id, upload_name, upload_root from tbl_upload WHERE upload_root = '"$page_id'"";
我无法获得正确的网页网址。我找不到正确的词语来描述我的问题。但我特别想要的是将page_id的确切值（自动递增）检索到upload_id中，如果不可能，则检索到upload _root。
我想做的是：
上传：
upload_id, upload_name, upload_root
(4, test, 4)
(5, test-sub, 4)
(6, test-sub2, 4)
<?php
if (isset($_GET['submit']))
{
$name = $_POST['title'];
$pageid = $_POST['page_id'];
$insert_sql = "INSERT INTO tbl_upload (upload_id, upload_name, upload_root) VALUES ('".$pageid."','".$name."','')";
mysql_query($insert_sql);
}
?>
<form action="your_page.php?submit=yes" enctype="multipart/form-data" method="post">
<div>Name: <input type="text" name="title" size="50"></div>
<input type="hidden" name="page_id" value= "<?php echo $page_id; ?>">
<input type="submit" value="Upload" id="upload" class="upload">
</form>
显然，插入页面的名称，上面写着"your_page.php"
您的代码容易受到SQL注入的攻击。请查阅如何为未来的项目实施PDO。