这个问题在这里已经有答案了:
我可以在php中混合mysql api吗(4个答案)
两年前关门了。
所以,我写了一个代码,管理员可以查看用户的用户名和密码。我写了一些代码,从w3schools复制了一些:p。所以。每当我运行代码时,它都会显示错误[调用资源上的成员函数query()]
这是我的密码:
进程.php
<?php
$username = $_POST['uname'];
$password = $_POST['pass'];
$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysql_escape_string($username);
$password = mysql_escape_string($password);
$conn = mysql_connect("localhost","root","");
mysql_select_db("Login3");
$result = mysql_query("select * from users where username = '$username' and password = '$password'");
$row = mysql_fetch_array($result);
if($row['username'] == $username && $row['password'] == $password) {
echo "Welcome " . $row['username'];
}
else {
echo "Invalid Credentials";
}
echo <h3>User List</h3>;
mysql_connect("localhost","root","");
mysql_select_db("Login3");
$sql = "SELECT id, username, password FROM users";
$request = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<br> id: ". $row["id"]. " - Name: ". $row["username"]. " " . $row["password"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
索引.html
<!DOCTYPE html>
<html>
<head>
<title>Login | Home</title>
</head>
<body>
<form action="process.php" method="POST">
<h3>Login</h3>
Username: <input type="text" name="uname"><br><br>
Password: <input type="password" name="pass"><br>
<input type="submit" name="btn">
</form>
<br>
<b>Test Accounts</b>
<table>
<tr>
<th>Username</th>
<th>Password</th>
</tr>
<tr>
<td>admin</td>
<td>admin@123</td>
</tr>
</table>
</body>
</html>
<style>
table, th, tr, td {
width: 25%;
border-collapse: collapse;
}
th, td {
border: 1px solid grey;
text-align: center;
}
tr {
background: white;
transition-duration: 0.3s;
}
tr:hover {
background: #ddd;
}
请更正我的密码,告诉我哪里错了。
1条答案
按热度按时间ctehm74n1#
首先,您应该创建一次性数据库连接,并遵循适当的查询语法。如果您想访问mysqli类的查询方法,那么您可以访问代码中不存在的mysqli类的查询方法,然后首先创建这个类的intance,然后编写适当的查询语法。我在下面更正了