查询mysql时出现OOP PHP错误

Question

我刚刚开始使用OOP，在使用教程时，我遇到了以下错误：

Warning: Missing argument 2 for MySQLDatabase::query(), called in
/Applications/MAMP/htdocs/object-oriented/public/index.php on line 7 and defined in
/Applications/MAMP/htdocs/object-oriented/includes/database.php on line 28

Notice: Undefined variable: sql in
/Applications/MAMP/htdocs/object-oriented/includes/database.php on line 29

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in
/Applications/MAMP/htdocs/object-oriented/includes/database.php on line 29

这是我的database.php

<?php

require('config.php');

class MySQLDatabase {

    public $connection;

    public function open_connection() {
        $this->$connection = mysqli_connect('DB_SERVER', 'DB_USER', 'DB_PASS');
        if (!$this->connection) {
            die("Database connection failed");
        } else {
            $db_select = mysqli_select_db($this->connection, DB_NAME);
            if (!$select_db) {
                die("Database selection failed");
            }
        }
    }

    public function close_connection() {
        if (isset($this->connection)) {
            mysqli_close($this->connection);
            unset($this->connection);
        }
    }

    public function query($connection, $sql) {
        $result = mysqli_query($this->connection, $sql);
        $this->confirm_query($result);
        return $result;
    }

以及我尝试在其中测试index.php ()方法的query。

$sql = "INSERT INTO users (id, username, password, first_name, last_name) ";
$sql .= "VALUES (1, 'andrei', 'password', 'Andrei', 'Popa')";
$result = $database->query($sql);

$sql = "SELECT * FROM users WHERE id = 1";
$result_set = $database->query($sql);
$found_user = mysqli_fetch_array($result_set);
echo "Found user!" . $found_user['username'];

知道我做错了什么吗？谢谢

Answer 1

public function query($connection, $sql) {

您不需要第一个参数，因为连接是您的类的属性。因此，只将$sql传递到查询方法中，就像您在使用该方法时所做的那样：

public function query($sql) {

Answer 2

这是你的定义：

public function query($connection, $sql) {

这是你的决定：

$result = $database->query($sql);

您需要添加一个连接:)

Answer 3

您有许多错误：

First

$db_select = mysqli_select_db($this->connection, DB_NAME);
if (!$select_db) {

检查变量db_select和select_db

第二个

public function query($connection, $sql) {

$result = $database->query($sql);

你需要检查你的论点。我认为，您需要从声明中删除$connection

public function query($sql) {

因为接下来您将使用类变量：

$this->connection;

Third

public function open_connection()

您的类中没有构造器，并且您没有调用方法open_connection，因此:您的活动连接是NULL

您需要将此方法添加到您的类中：

function __construct() {
    open_connection();
}