我想测试其中一位开发人员制作的登录功能。我想做的是在我的设置中创建测试用户，然后我想使用这些用户来测试登录功能。我希望看到当用户名和密码等于我们数据库中存储的值时它返回正确的布尔值。我的问题是，当我现在插入我的用户时，我以纯文本形式提供密码，因此它们以纯文本形式存储在数据库中，但是这里有一个问题。登录功能会在您尝试登录时对密码进行哈希处理，因为在注册完成后会对数据库中的密码进行哈希处理。所以基本上我刚刚插入的纯文本密码永远不会匹配，因为登录函数会散列密码以尝试找到匹配项。所以我需要做的是在我插入测试用户的密码时对其进行哈希处理。我该怎么做呢？这是我的代码目前的样子：
<?php
include 'functions.php';
class Test extends PHPUnit_Framework_TestCase {
protected function setUp(){
global $mysqli;
$mysqli = new mysqli('localhost', 'xxx', 'xxxxx', 'xxxxx');
$mysqli->query("INSERT INTO members (id, username, pnumber, password) VALUES ('200', 'testbrute', '920314', 'xxxxx')");
}
public function testLogin(){
global $mysqli;
$correctPass = Login('920314','xxxxxx', $mysqli);
$this->assertTrue($correctPass);
$wrongPass = Login('920314','xxxxxxxxx', $mysqli);
$this->assertFalse($wrongPass);
$NoUserExists = Login("980611-5298","--..--..--", $mysqli);
$this->assertFalse($NoUserExists);
}
protected function tearDown(){
$mysqli = new mysqli('localhost', 'xxx', 'xxxxx', 'xxxxx');
$mysqli->query("DELETE FROM members WHERE id IN (200)");
}
}
?>
这是登录功能的样子：
function login($pnumber, $password, $mysqli) {
// Using prepared Statements means that SQL injection is not possible.
if ($stmt = $mysqli->prepare("SELECT id, username, password, salt FROM members WHERE pnumber = ? LIMIT 1")) {
$stmt->bind_param('s', $pnumber); // Bind "$pnumber" to parameter.
$stmt->execute(); // Execute the prepared query.
$stmt->store_result();
$stmt->bind_result($user_id, $username, $db_password, $salt); // get variables from result.
$stmt->fetch();
$password = hash('sha512', $password.$salt); // hash the password with the unique salt.
if($stmt->num_rows == 1) { // If the user exists
// We check if the account is locked from too many login attempts
if(checkbrute($user_id, $mysqli) == true) {
// Account is locked
// Send an email to user saying their account is locked
return "account locked";
} else {
if($db_password == $password) { // Check if the password in the database matches the password the user submitted.
// Password is correct!
$ip_address = $_SERVER['REMOTE_ADDR']; // Get the IP address of the user.
$user_browser = $_SERVER['HTTP_USER_AGENT']; // Get the user-agent string of the user.
$user_id = preg_replace("/[^0-9]+/", "", $user_id); // XSS protection as we might print this value
$_SESSION['user_id'] = $user_id;
$username = preg_replace("/[^a-zA-Z0-9_\-]+/", "", $username); // XSS protection as we might print this value
$_SESSION['username'] = $username;
$_SESSION['login_string'] = hash('sha512', $password.$ip_address.$user_browser);
// Login successful.
return "login successful";
} else {
// Password is not correct
// We record this attempt in the database
$now = time();
$mysqli->query("INSERT INTO login_attempts (user_id, time) VALUES ('$user_id', '$now')");
return 'pass incorrect';
}
}
} else {
// No user exists.
return 'no exist';
}
}
}
我是 phpunit 和一般测试的新手，所以请过度描述。