php 从类中访问全局变量
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Access global variable from within a class
提问by pistacchio
I have the following (stripped down) code:
我有以下（精简）代码：
<?PHP
class A {
function Show(){
echo "ciao";
}
}
$a = new A();
$b = new B();
class B {
function __construct() {
$a->Show();
}
}
?>
With a bit of surprise I cannot access the globally defined $a variable from within the class and I get a Undefined variableexception. Any help?
有点惊讶的是，我无法从类中访问全局定义的 $a 变量，并且出现了未定义的变量异常。有什么帮助吗？
采纳答案by Franz
Why the surprise? That's a pretty logical variable scope problem there...
为什么会有惊喜？这是一个非常合乎逻辑的变量范围问题......
I suggest you use either the globalkeyword or the variable $GLOBALSto access your variable.
我建议您使用global关键字或变量$GLOBALS来访问您的变量。
EDIT:So, in your case that will be:
编辑：所以，在你的情况下，这将是：
global $a;
$a->Show();
or
或者
$GLOBALS['a']->Show();
EDIT 2:And, since Vinko is right, I suggest you take a look at PHP's manual about variable scope.
编辑 2：而且，由于 Vinko 是对的，我建议您查看 PHP 的有关变量范围的手册。
回答by Galen
please don't use the global method that is being suggested. That makes my stomach hurt.
请不要使用建议的全局方法。这让我的胃很痛。
Pass $a into the constructor of B.
将 $a 传递给 B 的构造函数。
class A {
function Show(){
echo "ciao";
}
}
$a = new A();
$b = new B( $a );
class B {
function __construct( $a ) {
$a->Show();
}
}
回答by LiraNuna
You will need to define it as a globalvariableinside the scope of the function you want to use it at.
您需要将其定义为要使用它的函数范围内的global变量。
function __construct() {
global $a;
$a->Show();
}
回答by Jay Zeng
<?php
class A {
public function Show(){
return "ciao";
}
}
class B {
function __construct() {
$a = new A();
echo $a->Show();
}
}
$b = new B();
?>
