laravel php pbkdf2登录验证

laravel php pbkdf2登录验证,php,mysql,laravel,hash,pbkdf2,Php,Mysql,Laravel,Hash,Pbkdf2,我有一个使用hash_pbkdf2的hash pasword的现有表。对于用户注册,它成功地插入mysql $string = mcrypt_create_iv(24, MCRYPT_DEV_URANDOM); $salt = strtoupper(bin2hex($string)); $hash = hash_pbkdf2("sha1", $data['password'], $string, 1000, 24, true); $hash = strtoupper(bin2hex($hash)

我有一个使用hash_pbkdf2的hash pasword的现有表。对于用户注册,它成功地插入mysql

$string = mcrypt_create_iv(24, MCRYPT_DEV_URANDOM);
$salt = strtoupper(bin2hex($string));
$hash = hash_pbkdf2("sha1", $data['password'], $string, 1000, 24, true);
$hash = strtoupper(bin2hex($hash));

return User::create([
    'name' => $data['name'],
    'email' => $data['email'],
    'hashedpassword' => $hash,
    'salt' => $salt,
]);
我无法使用它登录。这是我的密码

$found_salt = DB::table('users')->where('email', 'sicksand@gmail.com')->first();
$salt = $found_salt->salt;

echo "Salt : ".$salt."<br>";
$hash = hash_pbkdf2("sha1", "password", $salt, 1000, 24, true);
$hash = strtoupper(bin2hex($hash));

$userlogin = [
    'email' => "icksand@gmail.com",
    'hashedpassword' => $hash
];  
echo "Hash : ".$hash."<br>";

if(Auth::attempt($userlogin)) {
    echo "success";
} else {
    echo "not success";
}
$found\u salt=DB::table('users')->where('email','sicksand@gmail.com“)->first();
$salt=$found\u salt->salt;
回声“Salt:.$Salt.”
“; $hash=hash_pbkdf2(“sha1”,“password”,$salt,1000,24,true); $hash=strtoupper(bin2hex($hash)); $userlogin=[ '电子邮件'=>“icksand@gmail.com", 'hashedpassword'=>$hash ]; echo“Hash:.$Hash.”
“; if(Auth::trunt($userlogin)){ 呼应“成功”; }否则{ 呼应“不成功”; }

salt值相同,但哈希值不匹配。希望有人能帮忙。谢谢。

在您的第一个代码块中,您可以使用$string而不是$salt值对密码进行加密,但要将$salt存储到数据库中

因此,我认为您需要在第一个代码块中更改这一点:

$hash=hash_pbkdf2(“sha1”,$data['password'],$string,1000,24,true)

$hash=hash_pbkdf2(“sha1”,$data['password'],$salt,1000,24,true)