奇怪 ,$_POST[]取值报错
$name=$_POST['name'];
$password=$_POST['password'];
if ((!isset($name)) || (!isset($password))) {
?>
Please Log In
This page is secret.
} else {
$mysql = mysqli_connect("localhost", "webauth", "123456");
if(!$mysql) {
echo "Cannot connect to database.";
exit;
}
$selected = mysqli_select_db($mysql, "auth");
if(!$selected) {
echo "Cannot select database.";
exit;
}
$query = "select count(*) from authorised_users where
name = '".$name."' and
password = '".$password."'";
$result = mysqli_query($mysql, $query);
if(!$result) {
echo "Cannot run query.";
exit;
}
$row = mysqli_fetch_row($result);
$count = $row[0];
if ($count > 0) {
echo "
Here it is!
I bet you are glad you can see this secret page.
";
} else {
echo "
Go Away!
You are not authorized to use this resource.
";
}
}
?>
报错如下:Notice: Undefined index: name in C:\wamp\www\logindb.php on line 2
Notice: Undefined index: password in C:\wamp\www\logindb.php on line 3
以前貌似没这个问题,重新装了个版本的WAMP,出现了这个问题,奇怪
------解决方案--------------------不是错误
修改php.ini配置文件,error_reporting = E_ALL & ~E_NOTICE
------解决方案--------------------修改php.ini配置文件,error_reporting = E_ALL & ~E_NOTICE 问题解决了
这种只是屏蔽了 E_NOTICE的显示,实际问题还是存在的,就好像掩耳盗铃一样,问题出现了,但你假装不知道而已。
if(!empty($_POST['submit']))
{
$name=$_POST['name'];
$password=$_POST['password'];
}
这才是正确的方法,在使用POST数据之前,事先就要对数据的正确性进行判断,连input是否存在都不知道,就贸然使用,在生产环境是不可接受的。
------解决方案--------------------查看置顶的帖子里有基础问题里面说到你的问题了