Question

我的代码似乎有问题。如果没有cookie，它不会重定向，如果cookie存在，它不包括我的header.php。我制作了一个测试cookie脚本，我可以确认cookie是否存在。

代码：

//checks cookies to make sure they are logged in 
    if(isset($_COOKIE["user"]))
{
$username = $_COOKIE["user"]; 
$pass = $_COOKIE["password"];
$check = mysql_query("SELECT * FROM members WHERE email = '$username'") or die(mysql_error()); 
$info = mysql_fetch_array( $check );

//if the cookie has the wrong password, they are taken to the login page 
if ($pass != $info['password']) 
    {
    header("Location: login.php"); 
    exit();
    } 
else //otherwise they are shown the admin area   
    { 
    include 'header.php';
    }
}

Answer 1

除了脚本中的一堆不良做法之外，我认为导致故障的是：

$info = mysql_fetch_array( $check );
if ($pass != $info['password'])

如果你盯着手册，这就是它所说的

返回值

返回与获取的行对应的字符串数组，或   如果没有更多行，则为FALSE ...

所以代码看起来应该是

$info = mysql_fetch_array( $check );
if ($pass != $info[0]['password'])

或者在php gte 5.4中

$info = mysql_fetch_array( $check )[0];
if ($pass != $info['password'])

您应确保将display_errors设置为E_ALL，以便您可以更轻松地调试简单错误。

Answer 2

不推荐使用Mysql函数。应该使用Mysqli或PDO以及准备好的语句。示例：

//checks cookies to make sure they are logged in 
if(isset($_COOKIE["user"]) && isset($_COOKIE["password"])) {    
  $password = $_COOKIE["password"];
  /* Create the prepared statement */
  if ($stmt = $mysqli->prepare("SELECT password FROM members WHERE email = ?")) {

    /*bind parameters */
    $stmt->bind_param("s", $_COOKIE['user']);

    /* Execute the prepared Statement */
    $stmt->execute();

    /* Bind results to variables */
    $stmt->bind_result($dbPassword);

    /* fetch values */
    $stmt->fetch();

    if($password != $dbPassword) {
    //password didn't match, boot 'em
      header('Location:fail.php');
      exit();
    }
    //password matched, let 'em in
    else {
      header('Location:pass.php');
      exit();
    }
}
//no cookies set, boot 'em
else {
    header('Location:fail.php');
    exit();
}

附注：不要存储纯文本密码。您应该哈希密码并添加盐。

Cookie在PHP中无法正常工作以重定向

2 个答案: