如何使用PHP5类创建单例类?
当前回答
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 { }
其他回答
/**
* 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。
你可能应该添加一个私有的__clone()方法来禁止克隆实例。
private function __clone() {}
如果您不包含此方法,则可能出现以下情况
$inst1=UserFactory::Instance(); // to stick with the example provided above
$inst2=clone $inst1;
现在$inst1 !== $inst2 -它们不再是同一个实例了。
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();
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。
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。