4

我有一个非常简单的登录/用户注册脚本，它使用 sha1 和 salt 存储密码。我的密码和用户创建工作正常，并且将所有内容都存储在数据库中，但是当我尝试使用凭据登录时，它不起作用。搜索此主题时，我似乎找不到任何东西。

这是我的添加用户表单：

session_start();
include("includes/resume.config.php");

// make sure form fields have a value and strip them
function check_input($data, $problem='')
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
if ($problem && strlen($data) == 0)
{
    die($problem);
}
    return $data;
}

// get form values, escape them and apply the check_input function
$name = $link->real_escape_string(check_input($_POST['name'], "Please enter a name!"));
$email = $link->real_escape_string(check_input($_POST['email'], "Please enter an email!"));
$password = $link->real_escape_string(check_input($_POST['password'], "Please enter a password!"));

// generate a random salt for converting passwords into MD5
$salt = bin2hex(mcrypt_create_iv(32, MCRYPT_DEV_URANDOM));
$saltedPW =  $password . $salt;
$hashedPW = sha1($saltedPW);

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error());
// select the db
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name));

 // our sql query
$sql = "INSERT INTO admins (name, email, password, salt) VALUES ('$name', '$email', '$hashedPW', '$salt');";

//save the updated information to the database          
mysqli_query($link, $sql) or die("Error in Query: " . mysqli_error($link));

if (!mysqli_error($link)) 
{
    header("Location: file_insert.php");
}

这是我的登录脚本：这是行不通的 

function check_input($data, $problem='')
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
if ($problem && strlen($data) == 0)
{
    die($problem);
}
    return $data;
}

if(isset($_POST['submitLogin'])) { //form submitted?

// get form values, escape them and apply the check_input function
$name = $link->real_escape_string(check_input($_POST['name'], "Please enter a name!"));
$password = $link->real_escape_string(check_input($_POST['password'], "Please enter a password!"));

$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"');

$salt = mysqli_fetch_assoc($saltQuery);
$saltedPW =  $password . $salt;
$hashedPW = sha1($saltedPW);

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error());
// select the db
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name));

$validate_user = $link->query('SELECT id, name, password FROM admins WHERE name = "'.$name.'" AND password = "'.$hashedPW.'"');

if ($validate_user->num_rows == 1) {
    $row = $validate_user->fetch_assoc();
    $_SESSION['id'] = $row['id'];
    $_SESSION['loggedin'] = TRUE;
    Header('Location: file_insert.php');
} else {
    print "<center><p style='margin-top: 200px; font-weight: bold;'>Invalid Login Information</p>";
    print "<a href='admin-login.php'>Click here</a> to return to the login page.</center>";
}
}
4

2 回答 2

3

可能还会发生更多事情，但肯定它不起作用的一个原因是因为mysqli_fetch_assoc返回一个数组，并且您像字符串一样使用它。

当您调用时，PHP 会抱怨数组到字符串的转换，$password . $salt因为此时$salt是一个数组。结果是您将单词Array附加到密码中，从而导致哈希不正确。如果您已display_errors关闭和/或error_reporting设置隐藏通知，php.ini那么您将不会看到此消息。

如果你改变：

$saltedPW =  $password . $salt;

至：

$saltedPW =  $password . $salt['salt'];

那么它应该可以工作。

此外，您应该$salt在将其插入数据库之前转义，因为它可能包含空、不可打印或单/双引号，因为它是随机生成的。

于 2012-09-24T23:41:40.087 回答
0

您没有说是否收到任何错误消息，但从您发布的内容来看，我认为这是一个连接问题：

if(isset($_POST['submitLogin'])) { //form submitted?

// Here, you didn't connect to database, but you are expecting to fetch salt!
$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"');

$salt = mysqli_fetch_assoc($saltQuery);

因此，您可能必须先连接到数据库：

if(isset($_POST['submitLogin'])) { //form submitted?

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error());
// select the db
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name));

// Here, you didn't connect to database, but you are expecting to fetch salt!
$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"');

其次，mysqli_fetch_assoc 返回一个数组，其键名作为您在 SELECT 查询中嵌入的字段，最终代码应如下所示：

if(isset($_POST['submitLogin'])) { //form submitted?

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error());
// select the db
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name));

// Here, you didn't connect to database, but you are expecting to fetch salt!
$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"');
$salt = mysqli_fetch_assoc($saltQuery);
$saltedPW =  $password . $salt["salt"];
$hashedPW = sha1($saltedPW);
于 2012-09-24T23:47:15.167 回答