php 如何从同一个类的静态方法调用非静态方法？
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How to call non-static method from static method of same class?
提问by Rahul
I am working on PHPcode.
我正在处理PHP代码。
Here is the sample code to explain my problem:
这是解释我的问题的示例代码：
class Foo {
public function fun1() {
echo 'non-static';
}
public static function fun2() {
echo "static" ;
//self::fun1();
//Foo::fun1();
}
}
How can I call the non-static method from the static method ?
如何从静态方法调用非静态方法？
Note:Both functions are used throughout the site, which is not known. I can't make any changes in the static/non-static nature of them.
注意：这两个功能在整个站点中都使用，这是未知的。我无法对它们的静态/非静态性质进行任何更改。
回答by Mihai Matei
You must create a new object inside the static method to access non-static methods inside that class:
您必须在静态方法中创建一个新对象才能访问该类中的非静态方法：
class Foo {
public function fun1()
{
return 'non-static';
}
public static function fun2()
{
return (new self)->fun1();
}
}
echo Foo::fun2();
The result would be non-static
结果是 non-static
Later edit: As seen an interest in passing variables to the constructor I will post an updated version of the class:
稍后编辑：正如看到将变量传递给构造函数的兴趣，我将发布该类的更新版本：
class Foo {
private $foo;
private $bar;
public function __construct($foo, $bar)
{
$this->foo = $foo;
$this->bar = $bar;
}
public function fun1()
{
return $this->foo . ' - ' . $this->bar;
}
public static function fun2($foo, $bar)
{
return (new self($foo, $bar))->fun1();
}
}
echo Foo::fun2('foo', 'bar');
The result would be foo - bar
结果是 foo - bar
回答by optimoose
The main difference would be that you can call static methods for a class without having to instantiate an object of that class. So, in your static method try
主要区别在于您可以为类调用静态方法，而无需实例化该类的对象。所以，在你的静态方法中尝试
Foo $objInst = new Foo();
$objInst->fun1();
But I don't see how this would make any sense in any context.
但我不明白这在任何情况下都有什么意义。
回答by Fakhar Anwar
Asnwer selcted as correct solves problem. There is a valid use case (Design Pattern) where class with static member function needs to call non-static member function and before that this static members should also instantiate singleton using constructor a constructor.
Asnwer 选择正确解决问题。有一个有效的用例（设计模式），其中具有静态成员函数的类需要调用非静态成员函数，在此之前，该静态成员还应该使用构造函数和构造函数实例化单例。
Case:For example, I am implementing Swoole HTTP Request event providing it a call-back as a Class with static member. Static Member does two things; it creates Singleton Object of the class by doing initialization in class constructor, and second this static members does is to call a non-static method 'run()' to handle Request (by bridging with Phalcon). Hence, static class without constructor and non-static call will not work for me.
案例：例如，我正在实现 Swoole HTTP 请求事件，将其作为具有静态成员的类提供回调。静态成员做两件事；它通过在类构造函数中进行初始化来创建类的单例对象，第二个静态成员所做的是调用非静态方法“run()”来处理请求（通过与 Phalcon 桥接）。因此，没有构造函数和非静态调用的静态类对我不起作用。
