Question

我正在开发一个基本的登录系统，允许注册/登录将用户存储在MySQL数据库中。我有它所以人们可以注册，它存储用户名和散列密码在数据库中。当有人登录时，它会显示成功消息或错误消息，具体取决于用户名和密码是否匹配。

我的问题在于登录功能。我需要创建用户会话并将它们重定向到站点的仅登录部分，而不是显示当前存在的成功消息。我不确定如何做到这一点......

这是我的代码：

登录/注册功能

function login($username, $password) {
    $userpass = sha1($password);
    $result = mysqli_query($con, "SELECT * FROM members WHERE username='$username' AND password='$userpass'");
    while($row = mysqli_fetch_array($result)) {
        $success = true;
    }
    if($success == true) {
        echo 'Success!';
    } else {
        echo '<div class="alert alert-danger">Oops! It looks like your username and/or password are incorrect. Please try again.</div>';
    }
} // END LOGIN FUNCTION

function register($username, $password) {
    $userpass = sha1($password);

    // Check if Username Exists
    $result = mysqli_query($con,"SELECT * FROM members WHERE username='$username'");
    while($row = mysqli_fetch_array($result)) {
        $userexist = 1;
    }
    if($userexist > 0) {
        echo '<div class="alert alert-danger">Sorry, it looks like that username is already taken.</div>';
    } else {
        $newmember = "INSERT INTO members SET username='$username', password='$userpass'";
        if(mysqli_query($con,$newmember)) {
            echo '<div class="alert alert-success">Congrats! You can now log in using your username and password</div>';
        }
    }
}

Answer 1

这样想

function login($username, $password) {
    $userpass = sha1($password);
    $result = mysqli_query($con, "SELECT * FROM members WHERE username='$username' AND password='$userpass'");
    while($row = mysqli_fetch_array($result)) {
        $success = true;
    }
    if($success == true) {
        $_SESSION['username']= $username; 
        //redirect to home page
    } else {
        echo '<div class="alert alert-danger">Oops! It looks like your username and/or password are incorrect. Please try again.</div>';
    }
} // END LOGIN FUNCTION

Answer 2

这在很大程度上取决于您构建其余代码的方式。如果您只是在构建HTML页面并在其中添加PHP函数，那么您可以在代码的各个位置使用这么简单的东西：

$loggedIn = login($username, $password)

然后你需要让登录返回true或false。然后，您的路由结构可以根据$ loggedIn做出不同的决策，您的页面可以显示不同的布局和内容。

这在编程风格方面不是一种很好的风格，但它可能会给你一个想法。如果您添加有关如何构建代码的更多详细信息，则应获得更好的答案

Answer 3

   session_start();

    if($_SERVER["REQUEST_METHOD"] == "POST") {

$myusername=addslashes($_POST['username']);
$mypassword=addslashes($_POST['password']); 

    $sql="SELECT id FROM admin WHERE username='$myusername' and passcode='$mypassword'";
$result=mysql_query($sql);
    $row=mysql_fetch_array($result);


$active=$row['active']; 
    $count=mysql_num_rows($result); 
       if($count==1) {
              session_register("myusername"); 
                  $_SESSION['login_user']=$myusername;

              header("location: welcome.php"); 
         } 
       else { 
              $error="Your Login Name or
          Password is invalid";
            }
   }

Answer 4

为了给您Usman Allam答案一个变体，我认为简单地检查一下mysqli_num_rows（$ result）是否> = 1会容易得多

function login($username, $password) {
    $userpass = sha1($password);
    $result = mysqli_query($con, "SELECT * FROM members WHERE username='$username' AND password='$userpass'");
    if(mysqli_num_rows($result) >= 1) {
        $_SESSION['username'] = $username; 
        //redirect to home page
    } else {
        echo '<div class="alert alert-danger">Oops! It looks like your username and/or password are incorrect. Please try again.</div>';
    }
} // END LOGIN FUNCTION

如果用户名＆amp;密码正确，启动会话

4 个答案: