创建用户将信息保存到数据库中，但当我尝试签名时，只需自动登录，信息是否存储在数据库中。请帮忙。
这是我的server.php代码我认为问题出在这里，但我不确定。
<?php
session_start();
// initializing variables
$username = "";
$email = "";
$errors = array();
// connect to the database
$db = mysqli_connect('localhost', 'dbuser', 'dbpassword', 'dbname');
// REGISTER USER
if (isset($_POST['reg_user'])) {
// receive all input values from the form
$username = mysqli_real_escape_string($db, $_POST['username']);
$email = mysqli_real_escape_string($db, $_POST['email']);
$password_1 = mysqli_real_escape_string($db, $_POST['password']);
$password_2 = mysqli_real_escape_string($db, $_POST['password_2']);
// form validation: ensure that the form is correctly filled ...
// by adding (array_push()) corresponding error unto $errors array
if (empty($username)) { array_push($errors, "Username is required"); }
if (empty($email)) { array_push($errors, "Email is required"); }
if (empty($password_1)) { array_push($errors, "Password is required"); }
if ($password_1 != $password_2) {
array_push($errors, "The two passwords do not match");
}
// first check the database to make sure
// a user does not already exist with the same username and/or email
$user_check_query = "SELECT * FROM loginsystem WHERE username='$username' OR email='$email' LIMIT 1";
$result = mysqli_query($db, $user_check_query);
$user = mysqli_fetch_assoc($result);
if ($user) { // if user exists
if ($user['username'] === $username) {
array_push($errors, "Username already exists");
}
if ($user['email'] === $email) {
array_push($errors, "email already exists");
}
}
// Finally, register user if there are no errors in the form
if (count($errors) == 0) {
$password = md5($password_1);//encrypt the password before saving in the database
$query = "INSERT INTO loginsystem (username, email, password)
VALUES('$username', '$email', '$password')";
mysqli_query($db, $query);
$_SESSION['username'] = $username;
$_SESSION['success'] = "You are now logged in";
header('location: home.php');
}
}
// ...
// ...
// LOGIN USER
if (isset($_POST['login_user'])) {
$username = mysqli_real_escape_string($db, $_POST['username']);
$password = mysqli_real_escape_string($db, $_POST['password_1']);
if (empty($username)) {
array_push($errors, "Username is required");
}
if (empty($password)) {
array_push($errors, "Password is required");
}
if (count($errors) == 0) {
$password = md5($password);
$query = "SELECT * FROM users WHERE username='$username' AND password='$password'";
$results = mysqli_query($db, $query);
$_SESSION['username'] = $username;
$_SESSION['success'] = "You are now logged in";
header('location: home.php');
}else {
array_push($errors, "Wrong username/password combination");
}
}
?>
我的login.php是
<?php include('includes/server.php') ?>
<!DOCTYPE html>
<html>
<head>
<title>Clout-Cloud | LOGIN</title>
<link rel="stylesheet" type="text/css" href="css/style3.css">
</head>
<body>
<div class="container">
<section id="content">
<h1>CloutCloud Login</h1>
<div>
<form method="POST" action="login.php">
<div class="input-group">
<div class="input-group">
<label>Username</label>
<input type="text" name="username" required>
</div>
<div class="input-group">
<label>Password</label>
<input type="password" name="password_1" required>
</div>
<div class="input-group">
<button type="submit" class="button" name="login_user">Login</button>
</div>
<p>
<a href="recoverpassword.php">Lost your password?</a>
<a href="register.php">Register</a>
</div>
</div>
</div>
</form><!-- form -->
</div>
</section><!-- content -->
</div><!-- container -->
</body>
我没看到我真正出错的地方，但是有一个问题。请帮我解决。
更新 我尝试使用代码
从我的dabatase获取结果$password = md5($password);
$query = "INSERT INTO loginsyle (username, password)
VALUES('$username', $password')";
while($row = mysqli_fetch_assoc($result)) {
$username = $row['username'];
$password = $row['password'];
}
}else {
array_push($errors, "Wrong username/password combination");
}
尝试从我的查询中获取结果。虽然我认为其中的一部分已经解决了部分问题，即只能使用随机信息登录而没有注册表，但现在我得到了错误
Notice: Undefined variable: result in /home/u572108555/public_html/includes/server.php on line 72
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given in /home/u572108555/public_html/includes/server.php on line 72
可能导致这种情况的原因是什么？我试图以错误的方式从我的查询中获取结果吗？
答案 0 :(得分：0)
您没有验证查询的结果。 您应该检查返回的结果数= 1，小于（0）=用户详细信息不匹配。超过这个（2+）并且您遇到重叠登录凭据或错误查询的问题。
我还建议不要使用MD5作为加密
正如其他人所提到的，你很容易受到SQL注入攻击，使用条带斜线和转义来解决这个问题，比如
$myusername = stripslashes($myusername);
$myusername = mysqli_real_escape_string($myusername);
找出你可以使用像这样的函数的结果数
$sql="SELECT * FROM $tbl_name WHERE ID='$myusername' and password= '$password'";
$result=mysqli_query($sql);
// Mysqli_num_row is counting table row
$count=mysqli_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){
// user is logged in
}
答案 1 :(得分：-1)
您似乎无法验证$results = mysqli_query($db, $query)的结果。以下是代码，说明您需要做什么。
$results = mysqli_query($db, $query);
// Here you have to check the $result and see if you have one record selected.
// However you are not doing that currently.
$_SESSION['username'] = $username;
$_SESSION['success'] = "You are now logged in";
另一方面，你这样做的方式似乎有很多瑕疵。
server.php的文件具有所有例程。这将使您最终得到一个单片文件，这远不是可维护性。