我正在学习使用一些基于 mysqli 的视频教程制作网站。我开始知道使用准备好的语句更安全,我正在尝试创建一个登录系统。这是我到目前为止所做的。
此代码完全帮助我成功登录。
<form action ="" method="post">
User Name:<br/>
<input type='text' name='username' />
<br/><br/>
Password:<br/>
<input type='password' name='password' />
<br/><br/>
<input type='submit' name='submit' value='login'>
</form>
<?php
if(isset($_POST['submit'])){
$username = $_POST['username'];
$password = md5($_POST['password']);
$stmt = $con->prepare("SELECT username, password FROM users WHERE username=? AND password=? LIMIT 1");
$stmt->bind_param('ss', $username, $password);
$stmt->execute();
$stmt->bind_result($username, $password);
$stmt->store_result();
if($stmt->num_rows == 1) //To check if the row exists
{
while($stmt->fetch()) //fetching the contents of the row
{$_SESSION['Logged'] = 1;
$_SESSION['username'] = $username;
echo 'Success!';
exit();
}
}
else {
echo "INVALID USERNAME/PASSWORD Combination!";
}
$stmt->close();
}
else
{
}
$con->close();
?>
但我还需要检查用户是否未激活或已被禁止或停用。所以我又做了一个代码。
这是我编写的代码
<?php
if(isset($_POST['submit'])){
$username = $_POST['username'];
$password = md5($_POST['password']);
$stmt = $con->prepare("SELECT username, password FROM users WHERE username=? AND password=? LIMIT 1");
$stmt->bind_param('ss', $username, $password);
$stmt->execute();
$stmt->bind_result($username, $password);
$stmt->store_result();
if($stmt->num_rows == 1) //To check if the row exists
{
$result=$con->query($stmt);
$row=$result->fetch_array(MYSQLI_ASSOC);
$user_id= $row['user_id'];
$status = $row['status'];
if($status=='d'){
echo "YOUR account has been DEACTIVATED.";
}else{
$_SESSION['Logged'] = 1;
$_SESSION['user_id'] = $user_id;
$_SESSION['username'] = $username;
echo 'Success!';
exit();
}
}
else {
echo "INVALID USERNAME/PASSWORD Combination!";
}
$stmt->free_result();
$stmt->close();
}
else
{
}
$con->close();
?>
当我使用它时,出现以下错误
Warning: mysqli::query() expects parameter 1 to be string, object given in F:\XAMPP\htdocs\login\login.php on line 33
Fatal error: Call to a member function fetch_array() on a non-object in F:\XAMPP\htdocs\login\login.php on line 34
我有数据库表列
用户名, 用户名, 密码(md5), 用户级别, 状态。
在 user_level 下我有以下内容
a = admin
m = member
状态下
a = activated
n = not activated
d = deactivated
b = banned
登录时我需要检查用户状态,如果它被激活,它应该移动到索引页面,或者如果它是 d 它应该显示用户已经被停用,其他人也是如此。
如何在准备好的语句中做到这一点?
我在所有页面中都有这个 connect.php
?php
//error_reporting(0);
'session_start';
$con = new mysqli('localhost', 'username', 'password', 'database');
if($con->connect_errno > 0){
die('Sorry, We\'re experiencing some connection problems.');
}
?>
最佳答案
我认为您需要了解一下 mysqli_ 的工作原理。这应该会让您朝着正确的方向前进。
if(isset($_POST['submit'])){
$username = $_POST['username'];
$password = md5($_POST['password']);
$user_id = 0;
$status = ""
$stmt = $con->prepare("SELECT user_id, username, password, status FROM users WHERE username=? AND password=? LIMIT 1");
$stmt->bind_param('ss', $username, $password);
$stmt->execute();
$stmt->bind_result($user_id, $username, $password, $status);
$stmt->store_result();
if($stmt->num_rows == 1) //To check if the row exists
{
if($stmt->fetch()) //fetching the contents of the row
{
if ($status == 'd') {
echo "YOUR account has been DEACTIVATED.";
exit();
} else {
$_SESSION['Logged'] = 1;
$_SESSION['user_id'] = $user_id;
$_SESSION['username'] = $username;
echo 'Success!';
exit();
}
}
}
else {
echo "INVALID USERNAME/PASSWORD Combination!";
}
$stmt->close();
}
else
{
}
$con->close();
关于PHP MYSQLI 编写语句登录并查看用户状态,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17452614/