[PHP] 当一个类被另一个类使用,$this 所指是谁? - V2EX
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schezuk

[PHP] 当一个类被另一个类使用,$this 所指是谁?

  •  
  •   schezuk Feb 24, 2015 2934 views
    This topic created in 4083 days ago, the information mentioned may be changed or developed.

    提个问题,关于类的使用

    <?php
    class A {
    private $mysqli;
    public function foo(){
    echo get_class($this);
    $this->mysqli = new mysqli ( 'localhost', 'root', '', 'hdcms' );
    }
    }
    class B {
    function bar(){
    A::foo();// Note: this line will issue a warning if E_STRICT is enabled.
    }
    }
    $b = new B();
    $b->bar();
    ?>

    按官网说的,echo get_class($this);会输出'B'(PHP - 类与对象 - 基本概念
    那么$this->mysqli就是类B定义的$mysqli?

    如果类B没有定义$mysqli,会出现意外错误吗?
    或者类B需要定义$mysqli,那么会出现冲突吗?

    msg7086
        1
    msg7086  
       Feb 24, 2015
    A::foo()是static method call
    根据你引用的说明,「通常是该方法所从属的对象,但如果是从第二个对象静态调用时也可能是另一个对象」,所以这里的$this就是B,所以$this->mysqli就是B的mysqli。

    所以:
    会出错。
    不会冲突。
    schezuk
        2
    schezuk  
    OP
       Feb 24, 2015
    @msg7086 那么如何才能正确为类A定义私有对象$mysqli,使之不受A类以外的干扰,
    同时保证A类以内各方法都能安全使用这一对象?
    msg7086
        3
    msg7086  
       Feb 24, 2015   1
    首先你foo是定义为对象方法,mysqli也是对象属性,所以用A::foo()去调用本来就是有问题的。
    正确的做法有2种。
    1. 实例化A
    $a = new A()
    $a->foo();
    2. 定义foo为静态方法
    private static $mysqli;
    public static function foo() {
     self::$mysqli = new mysqli();
    }
    A::foo();

    如果你要让全局所有的函数都使用一份mysqli,你可以考虑使用单例模式。
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