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
我喜欢使用trait的@jose-segura方法,但不喜欢在子类上定义静态变量的需要。下面是一个解决方案,通过将实例缓存在一个静态局部变量中到按类名索引的工厂方法中来避免这种情况:
<?php
trait Singleton {
# Single point of entry for creating a new instance. For a given
# class always returns the same instance.
public static function instance(){
static $instances = array();
$class = get_called_class();
if( !isset($instances[$class]) ) $instances[$class] = new $class();
return $instances[$class];
}
# Kill traditional methods of creating new instances
protected function __clone() {}
protected function __construct() {}
}
用法与@jose-segura相同,只是在子类中不需要静态变量。
简单的例子:
final class Singleton
{
private static $instance = null;
private function __construct(){}
private function __clone(){}
private function __wakeup(){}
public static function get_instance()
{
if ( static::$instance === null ) {
static::$instance = new static();
}
return static::$instance;
}
}
希望有帮助。
不幸的是,当有多个子类时,Inwdr的答案就失效了。
下面是一个正确的可继承单例基类。
class Singleton
{
private static $instances = array();
protected function __construct() {}
protected function __clone() {}
public function __wakeup()
{
throw new Exception("Cannot unserialize singleton");
}
public static function getInstance()
{
$cls = get_called_class(); // late-static-bound class name
if (!isset(self::$instances[$cls])) {
self::$instances[$cls] = new static;
}
return self::$instances[$cls];
}
}
测试代码:
class Foo extends Singleton {}
class Bar extends Singleton {}
echo get_class(Foo::getInstance()) . "\n";
echo get_class(Bar::getInstance()) . "\n";
PHP 5.3允许通过后期静态绑定创建可继承的单例类:
class Singleton
{
protected static $instance = null;
protected function __construct()
{
//Thou shalt not construct that which is unconstructable!
}
protected function __clone()
{
//Me not like clones! Me smash clones!
}
public static function getInstance()
{
if (!isset(static::$instance)) {
static::$instance = new static;
}
return static::$instance;
}
}
这解决了一个问题,在PHP 5.3之前,任何扩展了Singleton的类都会生成父类的实例,而不是它自己的实例。
现在你可以做:
class Foobar extends Singleton {};
$foo = Foobar::getInstance();
$foo将是Foobar的一个实例而不是Singleton的一个实例。