Question

真是崩溃了..php代码出错求解

function login($username, $password)
{
$connect = db_connect_userauth();（这个函数经过注册步骤的验证已经没错）

$result = $connnect->query("select * from user //问题的line 8
where (username = '".$username."' or userID = '".$username."')
and password = sha1('".$password."')");

报错：
Notice: Use of undefined constant result - assumed 'result' in
D:\xampp\htdocs\test\php\db_connect.php on line
6

Notice: Undefined variable: connnect in
D:\xampp\htdocs\test\php\login_with_db.php on line
8

Fatal error: Call to a member function query() on a
non-object in D:\xampp\htdocs\test\php\login_with_db.php on line 8

求解.....真是找不到....
function db_connect_userauth()
{
$result = new mysqli('localhost', 'user', 'password', 'database');
if(!result)
{
throw new Exception（略写不下）
}
return $result;
}

Answer 1

在这个文件,去检查看下;

追问

这个文件没错了，只是一个notice不会影响运行，这个文件在另一个文件中已经成功运行了，文件如下：我写在补充里.....这打不下

追答

变量名写错了。。。