PHP类中的可变范围

Question

如何在此类中设置全局变量？我试过这个:

class myClass
{
   $test = "The Test Worked!";
   function example()
   {
      echo $test;
   }
   function example2()
   {
      echo $test." again";
   }
}

哪个无法加载页面完全引用500错误.接下来我尝试了这个:

class myClass
{
   public $test = "The Test Worked!";
   function example()
   {
      echo $test;
   }
   function example2()
   {
      echo $test." again";
   }
}

但是当我打印这两个时,我所看到的只是"再次"抱歉这么简单的问题!

谢谢!

Answer 1

这个变量可以像这样访问

echo $this->test;

Answer 2

如果需要实例变量(仅为该类的实例保留),请使用:

$this->test

(另有答案建议.)

如果你想要一个"class"变量,请在前面添加"static"关键字,如下所示:

类变量与实例变量不同,因为从类创建的所有对象实例将共享同一个变量.

(注意访问类变量,使用类名,或'self'后跟'::')

class myClass
{
   public static $test = "The Test Worked!";
   function example()
   {
      echo self::$test;
   }
   function example2()
   {
      echo self::$test." again";
   }
}

最后,如果你想要一个真正的常量(不可更改),在前面使用'const'(再次使用'self'加上'::'加上常量的名称(虽然这次省略'$'):

class myClass
{
   const test = "The Test Worked!";
   function example()
   {
      echo self::test;
   }
   function example2()
   {
      echo self::test." again";
   }
}

Answer 3

实际上有两种方法可以从类内或类外访问类中的变量或函数，如果它们请求项是公共的（或在某些情况下受保护）

class myClass
{
   public $test = "The Test Worked!";
   function example()
   {
      echo $this->test;
      // or with the scope resolution operator
      echo myClass::test;
   }
   function example2()
   {
      echo $this->test." again";
      // or with the scope resolution operator
      echo myClass::test." again";
   }
}

Answer 4

class Foo {

    public $bar = 'bar';

    function baz() {
        $bar;  // refers to local variable inside function, currently undefined

        $this->bar;  // refers to property $bar of $this object,
                     // i.e. the value 'bar'
    }
}

$foo = new Foo();
$foo->bar;  // refers to property $bar of object $foo, i.e. the value 'bar'

请从这里开始阅读:http://php.net/manual/en/language.oop5.basic.php