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php - 使用SQL检查数据库中是否存在用户

转载 作者:行者123 更新时间:2023-11-29 12:57:23 25 4
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我正在尝试检查名为 members 的表,以查看是否存在具有电子邮件和密码的用户。我能够连接到数据库,但由于某种原因,它会跳过所有这些 if 语句并回显 '您已登录!' 即使我输入电子邮箱或者密码错误?这是 html 和 php:

<form action="/login-user" method="POST">
Email: <input type="text" name="login_email"><br>
Password: <input type="password" name="login_password"><br>
<button type="submit">Login</button>
</form>

PHP:

<?php
session_start();
/*error_reporting(0);*/

require 'users/functions/user-functions.php';

require 'users/connect-database.php';

if (empty($_POST) === false) {
$email = mysqli_real_escape_string($connection, $_POST['login_email']);
$password = stripslashes(mysqli_real_escape_string($connection, $_POST['login_password']));
$encrypted_password = md5($password);
if (empty($email)) {
echo 'You need to enter an email<br>';
} else if (empty($password)) {
echo 'You need to enter a password<br>';
} else if(user_exists($connection, $email, $encrypted_password) === false) {
echo 'You don\'t seem to be registered?';
} else if (user_active($connection, $email, $encrypted_password) === false) {
echo 'You haven\'t activated your account!';
} else {
$login = login($connection, $email, $encrypted_password);
if ($login === false) {
echo 'That email/password combination is incorrect';
} else {
$_SESSION['user_id'] = $login;
$_SESSION['logged_in'] = true;
echo 'You have been logged in!';
}
}
/*print_r($errors);*/
} else {
echo 'inputs were empty<br>';
}
require 'users/disconnect-database.php';
?>

文件'user-functions.php'的内容:

<?php
function sanitize($connection, $data) {
return mysqli_real_escape_string($connection, $data);
}
function logged_in() {
return $_SESSION['logged_in'];
}
function user_exists($connection, $email, $password) {
$query = mysqli_num_rows(mysqli_query($connection, "SELECT * FROM members WHERE email = '$email' AND password = '$password'"));
return ($query > 0) ? true : false;
}
function user_active($connection, $email, $password) {
$query = mysqli_query($connection, "SELECT user_id FROM members WHERE email = '$email' AND password = '$password' AND `active` = 1");
return ($query !== false) ? true : false;
}
function return_user_id($connection, $email, $password) {
return mysqli_query($connection, "SELECT user_id FROM members WHERE email = '$email' AND password = '$password'");
}
function login($connection, $email, $password) {
/*$user_id = mysql_result(mysqli_query($connection, "SELECT user_id FROM members WHERE email = '$email' AND password = '$password'"), 0, 'user_id');*/
/*$password = md5($password);*/
$query = mysqli_query($connection, "SELECT user_id FROM members WHERE email = '$email' AND password = '$password'");
/*return (mysqli_query($connection, $query) or die (false));*/
if ($query === false) {
return false;
} else {
return $query;
}
/*return ($query !== false) ? true : false;*/
}
function log_out() {
unset($_SESSION['logged_in']);
session_unset();
session_destroy();
}
?>

如果答案是使用mysql_resultmysqli_result,请详细解释,因为即使在阅读了手册和W3Schools以及其他地方之后,我仍然不明白这些功能如何工作。

感谢您的回答,顺便说一句,我已经阅读了有关此内容的所有其他帖子,但我没有找到任何答案。谢谢。

最佳答案

首先,我真的建议使用 sha 来加密密码,因为 md5 很快就会被解密。

对于您的登录功能,请尝试以下操作:

<?php
function login($connection, $email, $password) {
$query = mysqli_query($connection, "SELECT `email`, `password` FROM `members` WHERE `email` = '$email' AND `password` = '$password'");
$count = mysqli_num_rows($query); //counting the number of returns

//if the $count = 1 or more return true else return false
if($count >= 1) {
return true;
} else {
return false;
}
}
?>

脚本返回 true 后,您可以设置 session 或执行您需要执行的操作。

编辑每个文件中都需要session_start,因此最好的办法就是包含它。我希望这能起作用,我输入得很快,所以可能会有一些错误,但请告诉我:

<?php
function generate($password) {
$password = hash('sha1', $password);
return $password;
}

function login($connection, $email, $password) {
$password = generate($password);

$query = mysqli_query($connection, "SELECT `email`, `password` FROM `members` WHERE `email` = '$email' AND `password` = '$password'");
$count = mysqli_num_rows($query); //counting the number of returns

//if the $count = 1 or more return true else return false
if($count >= 1) {
return true;
} else {
return false;
}
}

function exists($connection, $detail, $table, $row, $value) {
$query = mysqli_query($connection, "SELECT `$detail` FROM `$table` WHERE `$row` = '$value'");
$count = mysqli_num_rows($query);

if($count >= 1) {
return true;
} else {
return false;
}
}

function detail($connection, $detail, $table, $row, $value) {
$query = mysqli_query($connection, "SELECT `$detail` FROM `$table` WHERE `$row` = '$value'");
$associate = mysqli_fetch_assoc($query);

return $associate[$detail];
}

function errors($error) {
echo '<ul class="error">';
foreach($error as $fault) {
echo '<li>'.$fault.'<li>';
}
echo '</ul>';
}

function isLoggedIn() {
if(!empty($_SESSION['logged_in']) && exists($connection, 'id', 'members', 'id', $_SESSION['logged_in']) == true) {
return true;
} else {
return false;
}
}

function logout() {
unset($_SESSION['logged_in']);
}

if($_POST) {
$email = mysqli_real_escape_string($connect, strip_tags($_POST['email']));
$password = mysqli_real_escape_string($connect, strip_tags($_POST['password']));

if(!empty($email) && !empty($password)) {
if(!filter_var($email, FILTER_VALIDATE_EMAIL)) {
$error[] = 'Your email: '.$email.' is not valid';
}

if(exists($connection, 'email', 'members', 'email', $email) == false) {
$error[] = 'You are not registered';
}

if(detail($connection, 'active', 'members', 'email', $email) != 1) {
$error[] = 'Your account is not activated';
}

if(empty($error)) {
$query = login($connection, $email, $password);

if($query == true) {
$_SESSION['logged_in'] == detail($connection, 'id', 'members', 'email', $email);
}
}
} else {
$error[] = 'There are empty fields';
}

if(!empty($error)) {
echo errors($error);
}
}
?>

<form action="" method="POST">
Email: <input type="text" name="email"><br>
Password: <input type="password" name="password"><br>
<input type="submit" value="Login">
</form>

关于php - 使用SQL检查数据库中是否存在用户,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23834071/

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