accesing在PHP

Question

靜態變量我有類的簡單的情況下，靜態變量和get函數 所有編譯好，但在運行時我收到此錯誤

[Sun Jul 25 03:57:07 2010] [error] [client 127.0.0.1] PHP Fatal error: Undefined class constant 'TYPE' in ..... 

爲函數的getType（）

這裏是我的類

class NoSuchRequestHandler implements Handler{ 
    public static $TYPE = 2001; 
    public static $VER = 0; 

    public function getType(){ 
     return self::TYPE; 
    } 

    public function getVersion(){ 
     return self::VER; 
    } 
} 

謝謝大家

Answer 1

您可以訪問此兩種方式，因爲它是公共...

class NoSuchRequestHandler implements Handler{ 
    public static $TYPE = 2001; 
    public static $VER = 0; 

    public function getType(){ 
     return self::$TYPE; //not the "$" you were missing. 
    } 

    public function getVersion(){ 
     return self::$VER; 
    } 
} 

echo NoSuchRequestHandler::$TYPE; //outside of the class. 

Answer 2

PHP認爲你試圖訪問一個類常量，因爲：

return self::TYPE; 

http://www.php.net/manual/en/language.oop5.paamayim-nekudotayim.php

正如克里斯提到，使用：

return self::$TYPE;