php – 从sha1 salted密码验证用户登录

我有一个非常简单的登录/用户注册脚本,使用sha1和salt存储密码.我有密码和用户创建工作正常,并将所有内容存储在数据库中,但是当我尝试使用凭据登录时,它不起作用.在搜索这个主题时,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：这是不起作用的

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,"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_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>";
}
}

解决方法

可能会有更多的事情发生,但肯定的一个原因是它不起作用是因为 mysqli_fetch_assoc返回一个数组,而你正在使用它像一个字符串.

当你调用$password时,PHP会抱怨数组转换为字符串. $salt因为此时$salt是一个数组.结果是您在密码后面附加了单词Array,导致散列不正确.如果您关闭了display_errors和/或将error_reporting设置为隐藏PHP.ini中的通知,那么您将看不到此消息.

如果你改变：

$saltedPW =  $password . $salt;

至：

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

那它应该工作.

此外,您应该在将$插入数据库之前转义$salt,因为它可能包含null,unprintable或单/双引号,因为它是随机生成的.