Question

嗨，大家好我的第44行登录文件有问题（我会标记）。如果有人可以帮助我，我将不胜感激。它与第44行的else语句有关。谢谢

<?php
session_start();

$email = $_POST['email'];

$password = $_POST['password'];

if ($username&&$password){

$connect = mysql_connect("*****", "******", "*******") or die("Couldnt Connect to the Database");
mysql_select_db("1619882_members") or die("Couldnt Select/find the Database");

$query = mysql_query("SELECT * FROM social WHERE email='$email'");

$numrows = mysql_num_rows($query);

if($numrows!==0){
    while($row = mysql_fetch_assoc($query)){

        $dbusername = $row['email'];
        $dbpassword = $row['password'];
    }

    if($username==$dbusername&&sha1($password)==$dbpassword){

        echo "you are in";
        $_SESSION['email'] = $email;
    }
    else{
        echo "Your password is incorrect!";
        }

    (line 44) else{
        die("That user does not exist!");
            }


else
    die("Please enter a username and Password!!");  


?>

Answer 1

问题是你有2 else之后。

我知道你想在用户名错误和密码错误时给出单独的错误消息，所以你应该使用不同的IF。

我还假设您使用电子邮件地址登录。

if($username==$dbusername&&sha1($password)==$dbpassword){

    echo "you are in";
    $_SESSION['email'] = $email;
}
if($username!=$dbusername&&sha1($password)!=$dbpassword){
     echo "Password and username are wrong";
}
else{
    if($password!=$dbpassword){
        echo "Your password is incorrect!";
    }

    if($username!=$dbusername){
        die("That user does not exist!");
    }
}

登录脚本问题

1 个答案: