PHP 类函数内的全局变量
为什么在类内部的函数内部，我不能执行此语句：
global $connected = true;
但我可以这样做：
global $connected;
$connected = true;
How come inside a function which is inside a class, I can't do this statement:
global $connected = true;
But I can do this:
global $connected;
$connected = true;
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将
$connected引入作用域以及为其赋值是两件独立的事情。没有理由让它们在一份声明中成为可能，这实际上没有多大意义。
以下代码是否：
$x = 5引入作用域？5纳入作用域？5分配给全局$x？5分配给全局$x然后将$x纳入作用域？我当然知道你的意思是后者，而前两者没有任何意义。但是，拟议的声明并不清楚这一点。这将是糟糕的语法。
The bringing of
$connectedinto scope, and the assignment of a value to it, are two separate things.There is no reason for them to be possible in one statement, which wouldn't really make much sense.
Does the following code:
$x = 5into scope?5into scope?5to the global$x?5to the global$xand then bring$xinto scope?I know of course that you intend for it to mean the latter, and that the first two have no meaning. But, that is not clear from the proposed statement. It would be poor syntax.
因为在函数内部你首先必须声明一个全局变量。这是您必须在函数开始时执行的操作。这样你就可以激活某个未传递的变量。
Because inside a function you first have to announce a global variable. It is something you must do in the beginning of the function. That way you can activate a certain variable which was not passed through.