Question

我想显示：登录为： _ __ _ ___

到目前为止的代码：

登录检查，

<?php
include('config.php');

// username and password sent from form 
$myemail=$_POST['myemail']; 
$mypassword=$_POST['mypassword'];

// To protect MySQL injection
$myemail = stripslashes($myemail);
$mypassword = stripslashes($mypassword);
$myemail = mysql_real_escape_string($myemail);
$mypassword = mysql_real_escape_string($mypassword);

$sql="SELECT * FROM $tbl_name WHERE email='$myemail' and password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

if($count==1){
session_start(); 
$_SESSION['myemail'] = $myemail; 
header("Location: http://www.jblanksby.yourwebsolution.net/login_success.php?            user=$myemail"); } 
else 
{ header("Location: http://www.jblanksby.yourwebsolution.net/loginerror.php"); 
} 
?>

登录成功页面/第一个成员页面，

<? 
$email = $_GET['myemail']; 
session_start(); 
$_SESSION['myemail'] = $email;  

if(isset($_SESSION['email'])){ 
} else { 
echo " 
<script language='javascript'> 
alert('Sorry, but you must login to view the members area!') 
</script> 
<script> 
window.location='http://jblanksby.yourwebsolution.net/sign_in.php' 
</script> 
"; } 
?>

<html>
blah blah blah
</html>

用于显示用户电子邮件的代码

Logged in as: <? echo "$email"; ?>

事情的登录方面是完美的。仅显示用户电子邮件证明是困难的。我做错了什么/错过了什么？

Answer 1

由于您尝试在会话中显示值，因此必须使用$_SESSION['email']。

session_start(); // Or, if you have done it ahead, then omit this
echo "Logged in as:".$_SESSION['email'];

Answer 2

在$_SESSION上使用Starx建议，但您需要将session_start作为第一行PHP移动到页面上方$email = $_GET['myemail'];（或$email = $_SESSION['email']）

<?php
session_start();
$email = $_SESSION['email'];
?>

<强> ADDED

要使用会话变量，请将其声明为（假设您在表单提交时使用method = post）：

<?php
session_start();
$_SESSION['email'] = $_POST['email'];
?>

这当然对用户输入不安全，假设您在将用户输入从一个页面传递到另一个页面并将其注入查询之前对其进行了整理，这只是一个简单的例子。

Answer 3

登录脚本：

$username = $_POST['username'];
$password = $_POST['password'];
$password = sha1($password);

if ($username&&$password) {
        require_once('../db/nstDBconnector.php');
        $qr = "SELECT * FROM `users` WHERE `username` = '$username'";
        $query = mysql_query($qr);
        $numrows = mysql_num_rows($query);

        if ($numrows!=0) {
            // check password against the password in the database
            while ($row = mysql_fetch_array($query, MYSQL_ASSOC)) {
                $storedUsername = $row['username'];
                $storedPassword = $row['password'];
                $storedUserType = $row['user_type'];
                $storedUserFName = $row['full_name'];
                $_SESSION['last_ip'] = $_SERVER['REMOTE_ADDR'];
            }
            // check to see if username and password match
            if ($username==$storedUsername&&$password==$storedPassword) {
                ob_start();
                session_start();
                $_SESSION['username'] = $storedUsername;
                $_SESSION['user_type'] = $storedUserType;
                $_SESSION['full_name'] = $storedUserFName;
                header ('Location: main.php?login=successful');
            } else {
                header ('Location: index.php?error=incorrectPassword');
            }
        } else {
            header ('Location: index.php?error=noUser');
        }
    } else {
        header ('Location: index.php?error=missingUserAndPass');

    }

登录后您带他们去的页面：

if ($_SESSION['username']) {
    $userLogged = $_SESSION['username'];
    $userType = $_SESSION['user_type'];
    $userFName = $_SESSION['full_name'];
} else {
    header ('Location: http://www.yourdomain.com/index.php?error=notLoggedIn');
}

您要在页面上显示用户名的位置：

Logged in as <?php echo $userFName; ?>.

或

echo "Logged in as ", $userFName, ".";

我想在他们登录后显示在线用户

3 个答案: