检查会员状态和登录

有人可以推动我正确的方向，并告诉这个代码有什么问题。我没有收到任何错误，但它没有记录我。这个想法是在sql数据库中有一行叫memb，并检查状态是1还是0.如果它是1，它会把你带到受保护的页面，如果是0，它会给你subscibing页面检查会员状态和登录

<?php 


require_once('config.php'); 


if(isset($_POST['Login']) && isset($_POST['uname']) && isset($_POST['pass1'])) { 
// Here check if all input are sent 

$uname = $sql-> real_escape_string($_POST['uname']); 
$pass1 = $sql-> real_escape_string($_POST['pass1']); 


if (!empty($uname) && !empty($pass1)) { 
    // Check if you input are not empty 

    $query= mysqli_query($sql,"SELECT * FROM login WHERE uname='".$uname."' AND pass='".$pass1."' && memb='?'"); 
    if(!$query){ die(mysqli_error($sql)); } 
    $checkuser= mysqli_num_rows($query); 


    if($checkuser != 1) { 


    $error = "Username doesn't exist in our database!"; 
    } 

    // Change $login with $query 
    while ($row = mysqli_fetch_array($query)) { 
    $checkpass= $row['pass']; 

while ($row = mysqli_fetch_array($query)) { 
    $checkmemb= $row['memb']; 

    // here i changed $pass1 to $checkpass 
    if ($pass1 == $checkpass && $memb == 0){ 

    setcookie("user", $uname, time()+7200); 
    $_SESSION['user'] = $uname; 
    $_SESSION['start'] = time(); 
    $_SESSION['expire'] = $_SESSION['start'] + (60 * 60 * 60); 
    header("Location: pay.php"); 
    exit(); 
    } elseif($pass1 == $checkpass && $memb == 1) { 
    setcookie("user", $uname, time()+7200); 
    $_SESSION['user'] = $uname; 
    $_SESSION['start'] = time(); 
    $_SESSION['expire'] = $_SESSION['start'] + (60 * 60 * 60); 
    header("Location: main.html"); 
    exit(); 
    } 
    } 
} 
} 
}

来源

2013-08-18 user2694733

http://php.net/manual/en/function.error-reporting.php – VladHQ

尝试修改此：

if(isset($_POST['Login']) && isset($_POST['uname']) && isset($_POST['pass1'])) { 
// Here check if all input are sent

到：

if(isset($_POST['uname']) && isset($_POST['pass1'])) { 
// Here check if all input are sent

没有测试，这只是从快看猜测。但它似乎没有做任何事情，因为您的输入检查失败。

来源

2013-08-18 22:49:50 mrPowers

检查会员状态和登录

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