Question

EDITED

我有一个带有登录表单的Android项目。我也有一些用户存储在服务器的数据库中，我正在尝试创建一个连接，以便使用PHP文件登录。我的问题是，每次，即使我连接已经存在的用户我得到NT AUTHORITY\SYSTEM这是我尝试使用错误的凭据登录时应该采取的错误消息。它以某种方式传递了错误的参数。我有以下代码。

login.php

Login credentials are wrong. Please try again!

DB_Functions.php上的getUserByEmailAndPassword函数

<?php
require_once 'DB_Functions.php';
require_once 'DB_Connect.php';
$db = new DB_Functions();

// json response array
$response = array("error" => FALSE);

if (isset($_POST['email']) && isset($_POST['password'])) {

    // receiving the post params
    $email = $_POST['email'];
    $password = $_POST['password'];

    // get the user by email and password
    $user = $db->getUserByEmailAndPassword($email, $password);

    if ($user != false) {
        // user is found
        $response["error"] = FALSE;
        $response["oid"] = $user["oid"];
        $response["user"]["name"] = $user["name"];
        $response["user"]["surname"] = $user["surname"];
        $response["user"]["country"] = $user["country"];
        $response["user"]["email"] = $user["email"];
        $response["user"]["password"] = $user["password"];
        $response["user"]["salt"] = $user["salt"];
        $response["user"]["telephone"] = $user["telephone"];
        echo json_encode($response);
    } else {
        // user is not found with the credentials
        $response["error"] = TRUE;
        $response["error_msg"] = "Login credentials are wrong. Please try again!";
        echo json_encode($response);
    }
} else {
    // required post params are missing
    $response["error"] = TRUE;
    $response["error_msg"] = "Required parameters email or password are missing!";
    echo json_encode($response);
}
?>

我也使用base64加密。也许加密功能有问题。这是我的base64_encode函数，我在存储用户时使用它。它创建一个10位数的盐并将其存储到每个用户的DB中。盐的例子是public function getUserByEmailAndPassword($email, $password) { $stmt = $this->conn->prepare("SELECT * FROM owner WHERE email = ?"); $stmt->bind_param("s", $email); if ($stmt->execute()) { $stmt->bind_result($user['oid'], $user['name'], $user['surname'], $user['country'], $user['email'], $user['password'], $user['salt'], $user['telephone']); while ($stmt->fetch()) { //printf("%s %s\n", $email, $password); } $stmt->close(); // verifying user password $salt = $user['salt']; $encrypted_password = $user['password']; $hash = $this->checkhashSSHA($salt, $password); // check for password equality if ($encrypted_password == $hash) { // user authentication details are correct return $user; } } else { return NULL; } }。我知道盐的正确格式类似于2b67fd277b。我为什么要买这种盐？

cRDtpNCeBiql5KOQsKVyrA0sAiA=

解码

 public function hashSSHA($password) {

    $salt = sha1(rand());
    $salt = substr($salt, 0, 10);
    $encrypted = base64_encode(sha1($password . $salt, true) . $salt);
    $hash = array("salt" => $salt, "encrypted" => $encrypted);
    return $hash;
}

我在这里遗漏了什么吗？请帮忙，谢谢！

Answer 1

发现它！当我存储用户时，我使用的是$password而不是$encrypted_password，因此解密无法正确完成。最后在DB中，而不是密码列中的真实密码，有一个加密的字符串。

PHP：base64_encode没有解密正确编辑

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PHP：base64_encode没有解密正确*编辑*

1 个答案:

PHP：base64_encode没有解密正确编辑