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php - PHP 错误：当一切似乎都正确时，调用资源上的成员函数 query()

问题描述

所以，我只是写了一个代码，管理员可以查看用户的用户名和密码。我写了一些代码并从 w3schools 复制了一些：p。所以。每当我运行代码时，它都会显示错误 [Call to a member function query() on resource]

这是我的代码：

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;
    }

请更正我的代码并告诉我哪里错了。

标签： phpmysql

解决方案

首先，您应该创建一次数据库连接并遵循正确的查询语法。如果要访问查询方法，则访问代码中不存在的 mysqli 类的查询方法，然后首先创建此类的实例，否则编写正确的查询语法。我更正了请检查下面

<?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";
    }
?>
<h3>User List</h3>
<?php
    mysql_connect("localhost","root","");
    mysql_select_db("Login3");

    $sql = "SELECT id, username, password FROM users";
    $request = mysql_query($sql);
    if (mysql_num_rows($request) > 0) {
    while($row = mysql_fetch_array($request)) {
        echo "<br> id: ". $row["id"]. " - Name: ". $row["username"]. " " . $row["password"] . "<br>";
    }
} else {
    echo "0 results";
}

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