2010-05-06 73 views
0

我想要一个类属性引用另一个类,而不是它的对象,然后使用该属性来调用该类的静态方法。类组合而不是对象组合?

class Database { 
    private static $log; 

    public static function addLog($LogClass) { 
     self::$log = $LogClass; 
    } 

    public static function log() { 
     self::$log::write(); // seems not possible to write it like this 
    } 
} 

如何我可以做到这一点?

因为我没有理由让它们成为对象,所以我想为它使用类。

回答

1

既然你貌似只有在一个特定的方法/功能感兴趣(没有更多的合同/接口),你可以在一个方式,它不要紧写代码无论是静态方法还是对象方法(... hm,对象方法...听起来不对,什么是正确的名称...)或简单的函数。

class LogDummy { 
    public static function write($s) { 
    echo 'LogDummy::write: ', $s, "\n"; 
    } 
    public function writeMe($s) { 
    echo 'LogDummy->writeMe: ', $s, "\n"; 
    } 
} 

class Database { 
    private static $log=null; 

    public static function setLog($fnLog) { 
    self::$log = $fnLog; 
    } 

    public static function log($s) { 
    call_user_func_array(self::$log, array($s)); 
    } 
} 

// static method 
Database::setLog(array('LogDummy', 'write')); 
Database::log('foo'); 

// member method 
$l = new LogDummy; 
Database::setLog(array($l, 'writeMe')); 
Database::log('bar'); 

// plain old function 
function dummylog($s) { 
    echo 'dummylog: ', $s, "\n"; 
} 
Database::setLog('dummylog'); 
Database::log('baz'); 

// anonymous function 
Database::setLog(function($s) { 
    echo 'anonymous: ', $s, "\n"; 
}); 
Database::log('ham'); 

打印

LogDummy::write: foo 
LogDummy->writeMe: bar 
dummylog: baz 
anonymous: ham 
+0

对象的方法听起来不错,给我,但你可以把它叫做一个实例方法,也 – Gordon 2010-05-06 15:43:57

2

使用call_user_func功能:

class Logger { 
    public static function write($string) { 
     echo $string; 
    } 
} 

class Database { 
    private static $log; 

    public static function addLog($LogClass) { 
     self::$log = $LogClass; 
    } 

    public static function log($string) { 
     call_user_func(array(self::$log, 'write'), $string); 
    } 
} 

$db = new Database(); 
$db->addLog('Logger'); 
$db->log('Hello world!');