我正在学习用一些基于mysqli的视频教程制作网站。我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。以下是我迄今为止所做的工作。
这段代码帮助我完全登录成功。
<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
{
}
$con->close();
?>
但我也需要检查用户是否没有激活或被禁止或停用。所以我做了另一个代码。
这是我制作的代码
<?php
if(isset($_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();
?>
当我使用这个我得到以下错误
警告：mysqli:：query（）要求参数1为字符串，对象在F:''XAMPP''htdocs''login.php第33行中给出
致命错误：在第34行F:''XAMPP''htdocs''login.php中的非对象上调用成员函数fetch_array（）
我有数据库表列
user_ id，用户名，密码（md5），user_ level，地位
在user_level下，我有以下
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，或者如果它是d，它应该显示用户已经被停用，其他人也一样。
如何在事先准备好的陈述中做到这一点？
我在所有页面的中都有这个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.');
}
?>
我认为您需要了解mysqli_是如何工作的。这应该会让你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();