PHP代码不会使用"$ noteid"从数据库中删除项目.如果我在它的位置放一个数字,但是当我尝试使用"$ noteid"时.它不会这样做.它确实一切正确到它试图删除的程度.
这是我如何得到"$ noteid":
//javascript
function viewnote(noteid) {
window.location = "noteview.php?noteid=" + noteid;
}
//button in body
<input type="button" value="Edit" onclick="editnote('<?= $noteid ?>')" />
这是链接到页面的代码:
<?php
$noteid = $_REQUEST['noteid'];
if (isset($_POST['delete'])){
mysql_query("DELETE FROM notes WHERE noteid='$noteid'");
header ('Location: index2.php');
}
?>
<body>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" style="margin:0px; pading:0px"><input type="submit" name="delete" value="Delete"></form>
</body>
**现在正在工作!**使其发挥作用的是一个隐藏的形式领域.
这是代码:
<?php
if (isset($_POST['delete'])){
$nid = $_REQUEST['notenum'];
mysql_query("DELETE FROM notes WHERE noteid='$nid'");
header ('Location: index2.php');
}
?>
//body cody
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; …