我可以使用神奇的方法來捕捉Access to undeclared static property不可訪問的靜態屬性嗎? 例如, class greeting
{
static public function init()
{
static::$message = 'Hello World!';
}
/*
* Set the ina
我想定義一個可以迭代的單個對象,而不必創建一個類,然後創建一個實例。像這樣的東西: class Thing(object):
stuff = ["foo", "bar", "baz"]
@classmethod
def __iter__(cls):
return iter(cls.stuff)
for thing in Thing:
pri
類 獲取屬性調用我有以下代碼: class Page extends APage{
/**
* @findby selector
*/
protected $property
public function getPropertyCount(){
count($this->property);
}
}
class
我的代碼有問題,它始終有錯誤'file' object has no attribute '__getitem__'。這裏是我的代碼: def passHack():
import random
p1 = random.choice(["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r",
我試圖重新定義特徵中魔法__set的行爲。當我還想從特徵訪問父類定製__set函數時出現問題。直到我也需要使用主類__set方法,因爲它在做一些其他的東西與變量集 trait TestingTrait {
public function __set($key, $value)
{
// Some stuff...
parent::__set($ke
我有一個類應該用作字符串,並且將始終是一個字符串(即使爲空)。該對象將始終具有字符串表示形式。以下是我班的一個示例: class Something
def initialize
@real_string_value = "hello"
end
def to_s
return @real_string_value
end
d