PHPUnit如何mock函数内部实例化的对象

[m718281962]

class cash{
    public function __construct(){}
    public function isCashed(){
       //do something
        return 1;
    }
}

class sendMsg{
    public function send(){
          $cash = new cash();
          $isCashed = $cash->isCashed();
          return $isCashed;
         //do something
    }
}

我想要对上面的sendMsg进行测试的时候 请问怎样mock出cash类然后注入到sendMsg当中

class sendMsgTest extension PHPUnit_Framework_TestCase{
      
       public function testSend(){
             $sendMsg = new sendMsg();
             
             //如何mock一个cash
      }
}

[碼上道]

class cash 这个已经定义好了吗，还是没有定义？

[m718281962]

引用 1 楼 jerry_dqh 的回复:

class cash 这个已经定义好了吗，还是没有定义？

都已经定义好了，是对现有代码写测试，上面那个只是我简化了的个东西

[碼上道]

可以试试phpunit phpmock

[m718281962]

在《修改代码的艺术》一书中找到了答案:

class sendMsg{
    protected function cash(){
         return new cash()
    }
    public function send(){
          $cash = $this->cash();
          $isCashed = $cash->isCashed();
          return $isCashed;
         //do something
    }
}
//将外部依赖抽离到单独的函数 然后测试的时候创建一个子类
class sendMsgExt extend sendMsg{
        protected $cash;
        public function setCash($cash){
                return $this->cash = $cash;
        }
        protected function cash(){
            return $this->cash;
        }
}
//测试的时候 实例化子类并通过setCash将mock出来的cash注入进来
class sendMsgTest extension PHPUnit_Framework_TestCase{
       
       public function testSend(){
             $sendMsg = new sendMsgExt();
             $mockCash = $this->mock('cash');
             $mockCash->expects($this->any())->method("isCashed")->will($this->returnValue(1));
              $sendMsg->setCash($mockCash);
              $this->assertEquals(1,$sendMsg->send());
      }
}