如何使用PHP5类创建单例类?
当前回答
以上答案都是可以的,但我还会再补充一些。
无论谁在2021年来到这里,我都将展示另一个使用单例模式类作为trait的例子,并在任何类中重用它。
<?php
namespace Akash;
trait Singleton
{
/**
* Singleton Instance
*
* @var Singleton
*/
private static $instance;
/**
* Private Constructor
*
* We can't use the constructor to create an instance of the class
*
* @return void
*/
private function __construct()
{
// Don't do anything, we don't want to be initialized
}
/**
* Get the singleton instance
*
* @return Singleton
*/
public static function getInstance()
{
if (!isset(self::$instance)) {
self::$instance = new self();
}
return self::$instance;
}
/**
* Private clone method to prevent cloning of the instance of the
* Singleton instance.
*
* @return void
*/
private function __clone()
{
// Don't do anything, we don't want to be cloned
}
/**
* Private unserialize method to prevent unserializing of the Singleton
* instance.
*
* @return void
*/
private function __wakeup()
{
// Don't do anything, we don't want to be unserialized
}
}
所以,像在任何课堂上一样轻松地使用它。假设,我们想在UserSeeder类中实现单例模式。
<?php
class UserSeeder
{
use Singleton;
/**
* Seed Users
*
* @return void
*/
public function seed()
{
echo 'Seeding...';
}
}
其他回答
这是在数据库类上创建单例的例子
设计模式 1)单
class Database{
public static $instance;
public static function getInstance(){
if(!isset(Database::$instance)){
Database::$instance=new Database();
return Database::$instance;
}
}
$db=Database::getInstance();
$db2=Database::getInstance();
$db3=Database::getInstance();
var_dump($db);
var_dump($db2);
var_dump($db3);
那么输出是
object(Database)[1]
object(Database)[1]
object(Database)[1]
只使用单个实例,不要创建3个实例
class Database{
//variable to hold db connection
private $db;
//note we used static variable,beacuse an instance cannot be used to refer this
public static $instance;
//note constructor is private so that classcannot be instantiated
private function __construct(){
//code connect to database
}
//to prevent loop hole in PHP so that the class cannot be cloned
private function __clone() {}
//used static function so that, this can be called from other classes
public static function getInstance(){
if( !(self::$instance instanceof self) ){
self::$instance = new self();
}
return self::$instance;
}
public function query($sql){
//code to run the query
}
}
Access the method getInstance using
$db = Singleton::getInstance();
$db->query();
protected static $_instance;
public static function getInstance()
{
if(is_null(self::$_instance))
{
self::$_instance = new self();
}
return self::$_instance;
}
这段代码可以应用于任何类,而无需关心类名。
/**
* Singleton class
*
*/
final class UserFactory
{
private static $inst = null;
// Prevent cloning and de-serializing
private function __clone(){}
private function __wakeup(){}
/**
* Call this method to get singleton
*
* @return UserFactory
*/
public static function Instance()
{
if ($inst === null) {
$inst = new UserFactory();
}
return $inst;
}
/**
* Private ctor so nobody else can instantiate it
*
*/
private function __construct()
{
}
}
使用方法:
$fact = UserFactory::Instance();
$fact2 = UserFactory::Instance();
$fact == $fact2;
But:
$fact = new UserFactory()
抛出错误。
参见http://php.net/manual/en/language.variables.scope.php#language.variables.scope.static了解静态变量的作用域以及为什么设置static $inst = null;的工作原理。
我同意第一个答案,但我也会将类声明为final,这样它就不能被扩展,因为扩展单例违背了单例模式。此外,实例变量应该是私有的,这样就不能直接访问它。还要将__clone方法设为私有,这样就不能克隆单例对象。
下面是一些示例代码。
/**
* Singleton class
*
*/
final class UserFactory
{
private static $_instance = null;
/**
* Private constructor
*
*/
private function __construct() {}
/**
* Private clone method
*
*/
private function __clone() {}
/**
* Call this method to get singleton
*
* @return UserFactory
*/
public static function getInstance()
{
if (self::$_instance === null) {
self::$_instance = new UserFactory();
}
return self::$_instance;
}
}
示例使用
$user_factory = UserFactory::getInstance();
这阻止你做什么(这将违反单例模式..
你不能这么做!
$user_factory = UserFactory::$_instance;
class SecondUserFactory extends UserFactory { }