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 $this->Run("SELECT * FROM categories");
}
}
现在的问题是,当我通过创建基类的对象来运行Run()方法时,它工作正常.但是当我创建对象对象时,子类ie类和执行方法viewCategories(); 我收到以下错误
警告:缺少Database :: __ construct()的参数1,在第16行的E:\ xampplite\htdocs\ecommerce\index.php中调用,并在E:\ xampplite\htdocs\ecommerce\classes\class.database.php中定义第17行
警告:缺少Database :: __ construct()的参数2,在第16行的E:\ xampplite\htdocs\ecommerce\index.php中调用,并在E:\ xampplite\htdocs\ecommerce\classes\class.database.php中定义第17行
警告:缺少Database :: __ construct()的参数3,在第16行的E:\ xampplite\htdocs\ecommerce\index.php中调用,并在E:\ xampplite\htdocs\ecommerce\classes\class.database.php中定义第17行
警告:缺少Database :: __ construct()的参数4,在第16行的E:\ xampplite\htdocs\ecommerce\index.php中调用,并在E:\ xampplite\htdocs\ecommerce\classes\class.database.php中定义第17行
警告:mysqli_query()期望参数1为mysqli,在第35行的E:\ xampplite\htdocs\ecommerce\classes\class.database.php中给出null.无法执行请求
Udated:这就是我调用方法的方式
<?php
function __autoload($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.
您不应该为has-a关系使用继承,继承描述is-a关系.在你的情况下,Categories 不是数据库,Categories 有一个数据库.
使用组合代替:
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);