dongnachuang6635 2014-08-28 07:58
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已采纳

Phpunit mock没有效果

code:

class MockMe
{
    public function mockMeee()
    {
        return 'Im not mocked';
    }
}

test:

$sut = new MockMe();
$this
    ->getMock(get_class($sut))
    ->expects($this->any())
    ->method('mockMeee')
   ->will($this->returnValue('Im finally mocked'));
echo $sut->mockMeee();

this outputs the original "Im not mocked", but it supposed to send the Im finally mocked text. What is wrong?

EDIT: done:

$stub = $this->getMock('MockMe');
$stub->method('mockMeee')->willReturn('Im finally mocked');
echo $stub->mockMeee();
  • 写回答

1条回答 默认 最新

  • douwei1904 2014-08-28 08:04
    关注

    You are constructing the real MockMe, then building a mock that you are doing nothing with. I think your test should be something like:

    $sut = $this->getMock('MockMe');
    
    $sut->expects($this->any())
        ->method('mockMeee')
        ->will($this->returnValue('Im finally mocked'));
    
    echo $sut->mockMeee();
    

    Refer to http://phpunit.de/manual/4.2/en/test-doubles.html for more information on PHPUnit mocks.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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