laravel PHPUnit 和来自 Guzzle 的模拟请求
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PHPUnit and mock request from Guzzle
提问by themazz
I have a class with the following function :
我有一个具有以下功能的类：
public function get(string $uri) : stdClass
{
$this->client = new Client;
$response = $this->client->request(
'GET',
$uri,
$this->headers
);
return json_decode($response->getBody());
}
How can I mock the request method from PHPUnit? I tried different ways but it always tries to connect to the uri specified.
如何从 PHPUnit 模拟请求方法？我尝试了不同的方法，但它总是尝试连接到指定的 uri。
I tried with :
我试过：
$clientMock = $this->getMockBuilder('GuzzleHttp\Client')
->setMethods('request')
->getMock();
$clientMock->expects($this->once())
->method('request')
->willReturn('{}');
But this didn't work. What can I do? I just need to mock the response to be empty.
但这没有用。我能做什么？我只需要模拟响应为空。
Thanks
谢谢
PD : Client comes from (use GuzzleHttp\Client)
PD ：客户端来自（使用 GuzzleHttp\Client）
回答by themazz
I think as suggested is better to use http://docs.guzzlephp.org/en/stable/testing.html#mock-handler
我认为建议最好使用http://docs.guzzlephp.org/en/stable/testing.html#mock-handler
as it looks like the most elegant way to do it properly.
因为它看起来是最优雅的正确方式。
Thank you all
谢谢你们
回答by duncan
The mocked Response doesn't need to be anything in particular, your code just expects it to be an object with a getBodymethod. So you can just use a stdClass, with a getBodymethod which returns some json_encodedobject. Something like:
模拟的 Response 不需要是任何特别的东西，您的代码只希望它是一个带有getBody方法的对象。所以你可以只使用stdClass, 和一个getBody返回一些json_encoded对象的方法。就像是：
$jsonObject = json_encode(['foo']);
$uri = '/foo/bar/';
$mockResponse = $this->getMockBuilder(\stdClass::class)->getMock();
mockResponse->method('getBody')->willReturn($jsonObject);
$clientMock = $this->getMockBuilder('GuzzleHttp\Client')->getMock();
$clientMock->expects($this->once())
->method('request')
->with(
'GET',
$uri,
$this->anything()
)
->willReturn($mockResponse);
$result = $yourClass->get($uri);
$expected = json_decode($jsonObject);
$this->assertSame($expected, $result);
