Question

下面是我的基类，即数据库方法。

// Constructor
public function __construct($argHost, $argUsername, $argPassword, $argDatabase)
{
    $this->_host = $argHost;
    $this->_username = $argUsername;
    $this->_password = $argPassword;
    $this->_database = $argDatabase;
}

// Connect to the database
public function Connect()
{
    if (!$this->Is_Connected()) {
        $this->_connection = mysqli_connect($this->_host,$this->_username,$this->_password,$this->_database);    
    } else {
        return $this->_connection;
    }

}
// Run query
public function Run($query)
{
    if ($this->result = mysqli_query($this->_connection,$query)) {
        return $this->result;
    } else {
        die("Couldn't perform the request");
    }
}

我的孩子课是下面的分类方法

class Categories extends Database
{    
    public $category_id = '';
    public $category_name = '';
    public $category_image = '';

    // View Category
    public function ViewCategories() 
    {
        return $this->Run("SELECT * FROM categories");
    }       
}

现在的问题是，当我通过创建基类的对象来运行Run（）方法时，它工作正常。但是当我创建对象对象时，子类即类别并执行方法viewCategories（）;我收到以下错误

警告：缺少参数1，用于调用Database :: __ construct（）   E：\ xampplite \ htdocs \ ecommerce \ index.php在第16行并在中定义   第17行的E：\ xampplite \ htdocs \ ecommerce \ classes \ class.database.php

警告：缺少数据库:: __ construct（）的参数2，调用   E：\ xampplite \ htdocs \ ecommerce \ index.php在第16行并在中定义   第17行的E：\ xampplite \ htdocs \ ecommerce \ classes \ class.database.php

警告：缺少数据库:: __ construct（）的参数3，调用   E：\ xampplite \ htdocs \ ecommerce \ index.php在第16行并在中定义   第17行的E：\ xampplite \ htdocs \ ecommerce \ classes \ class.database.php

警告：缺少数据库:: __ construct（）的参数4，调用   E：\ xampplite \ htdocs \ ecommerce \ index.php在第16行并在中定义   第17行的E：\ xampplite \ htdocs \ ecommerce \ classes \ class.database.php

警告：mysqli_query（）期望参数1为mysqli，null给定   在第35行的E：\ xampplite \ htdocs \ ecommerce \ classes \ class.database.php中   无法执行请求

Udated：这就是我调用方法的方法

<?php
function __autoload($class_name) {
    include 'classes/class.'.$class_name . '.php';
}
$connection = new Database("localhost", "raheel", "raheel786", "ecommerce");
$connection->Connect();
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Ecommerce</title>
</head>
<body>
   <?php 
   $category = new Categories();
   $category_list = $category->ViewCategories();
   var_dump($category_list);
   ?> 
</body>
</html>

请帮我解决这个问题。

Answer 1

您不应该为has-a relationship使用继承，继承描述is-a relationship。在您的情况下，类别不是数据库，类别具有数据库。

改为使用构图：

class Categories
{
    private $database;

    function __construct(Database $database)
    {
        $this->database = $database;
    }

    public function ViewCategories()
    {
        return $this->database->Run("SELECT * FROM categories");
    }
}

用法：

$connection = new Database("localhost", "raheel", "raheel786", "ecommerce");
$connection->Connect();
// ...
$category = new Categories($connection);
$category_list = $category->ViewCategories();
var_dump($category_list);

Answer 2

在您的情况下，您需要重载构造函数。只需将空__construct()方法添加到子类

即可

class Categories extends Database{

    public function __construct() {}

    ...

}

但它无法解决您的问题：）

作为一种解决方案，我可以看到两种变体：

1）制作静态变量_connection，使其可用于所有对象

public function Connect(){
    if( ! self::$_connection){
        self::$_connection = mysqli_connect($this->_host,$this->_username,$this->_password,$this->_database);    
    }

    return self::$_connection;
}

public function Run($query){
    if($this->result = mysqli_query(self::$_connection,$query)){
    return $this->result;
                }
                else 
                   die("Couldn't perform the request");
}

2）我认为这是更好的方式。使用连接和查询创建2个独立的类，以便包含另一个

class Database {
    // Constructor
    public function __construct($argHost,$argUsername,$argPassword,$argDatabase){
        $this->_host = $argHost;
        $this->_username = $argUsername;
        $this->_password = $argPassword;
        $this->_database = $argDatabase;
    }

    // Connect to the database
    public function Connect(){
        if(!$this->Is_Connected()){
            $this->_connection = mysqli_connect($this->_host,$this->_username,$this->_password,$this->_database);    
        } else {
            return $this->_connection;
        }
    }
    // Run query
    public function Run($query){
        if($this->result = mysqli_query($this->_connection,$query)){
            return $this->result;
        }
        else 
           die("Couldn't perform the request");
    }
}


class Categories extends Database{
    private $db;

    public function __construct(Database $db) {
        $this->db = $db;
    }

    public $category_id = '';
    public $category_name = '';
    public $category_image = '';

    // View Category
    public function ViewCategories() {
        return $this->db->Run("SELECT * FROM categories");
    }   
}

所以用法是：

$connection = new Database("localhost", "raheel", "raheel786", "ecommerce");
$connection->Connect();

$category = new Categories($connection);

注意！我没有测试它，只是举个例子

如何在子类中调用基类的构造函数

2 个答案: