Question

如果网站正在维护中，我想在登录时输出一条错误消息，但是我当前的代码无法正常工作，并且似乎只是在运行维护代码而已。我想这样，如果我已经定义为$maintenance的MySQL数据库中的维护列为空，则用户可以像往常一样登录，但是如果包含1，则用户将看到错误消息，但是阵列中具有其IP的管理员仍可以登录。我已经在$maintenance中已经包含的另一个文件中定义了class.user.php。代码在下面。

Settings.php

$auth_user = new USER();

$site_name = $auth_user->runQuery("SELECT * FROM `settings` LIMIT 1");
$site_name->execute();
while ($show = $site_name -> fetch(PDO::FETCH_ASSOC)){
        $maintenance = $show['maintenance'];
}

Class.user.php

require_once('settings.php');

....other functions here
....other functions here
.....other functions here
.....

        public function doLogin($uname,$umail,$upass)
    {
        try
        {
            $stmt = $this->conn->prepare("SELECT user_id, user_name, user_email, user_pass, status FROM users WHERE user_name=:uname OR user_email=:umail ");
            $stmt->execute(array(':uname'=>$uname, ':umail'=>$umail));
            $userRow=$stmt->fetch(PDO::FETCH_ASSOC);
            if($stmt->rowCount() == 1) 
            {
                if(password_verify($upass, $userRow['user_pass']))
                {
                    session_regenerate_id(false);
                    return ["correctPass"=>true, "banned"=> ($userRow['status']== 1) ? true : false, "maintenance"=> ($maintenance== 1) ? true : false];

                }
                else
                {
                    return ["correctPass"=>false];
                }
            }
        }
        catch(PDOException $e)
        {
            echo $e->getMessage();
        }
    }

Login.php

$validation = $login->doLogin($uname,$umail,$upass);
if($validation["correctPass"]){
    if($validation["maintenance"]){
        if (!in_array(@$_SERVER['REMOTE_ADDR'], array('1.1.1.1'))){
            $error = "Website under maintenance";
        }
    }
    if($validation["banned"]){
        $error = "User has been banned";
    }else{
        if(Token::check($_POST['token'])) {
        $stmtt = $login->runQuery("SELECT user_id FROM users WHERE user_name=:uname OR user_email=:umail ");
        $stmtt->execute(array(':uname'=>$uname, ':umail'=>$umail));
        $userRow=$stmtt->fetch(PDO::FETCH_ASSOC);
        $_SESSION['user_session'] = $userRow['user_id'];
        $success = "Logged in successfully, redirecting..";
        header( "refresh:3;url=dashboard" );
        } else {
            $error = "Unexpected error occured";
        }
    }
}
else{
    $error = "Incorrect username/email or password";
}

Answer 1

正如其他人在评论中指出的那样，$maintenance在您的doLogin函数的范围之外。如果您只想将其用作全局变量，可以像下面这样设置doLogin函数：

public function doLogin($uname,$umail,$upass)
{
    global $maintenance;
    ...

使用global关键字可以访问当前函数范围之外的变量。更好的方法可能是将$maintenance变量作为如下参数传递给函数：

public function doLogin($uname,$umail,$upass,$maintenance)
{
    ...

然后仅在Login.php文件中使用，就像这样：

$validation = $login->doLogin($uname,$umail,$upass,$maintenance);

这些选项对您有用吗？

检查网站是否正在维护中

1 个答案: