如何为MockHttpClient创建一个具有statusCode 400标头的MockResponse

Question

我想使用symfony/http-client测试API的不同响应，但如何创建statusCode 400？

这就是我尝试的：

public static function mockShipmentResponseFail(): MockResponse
{
$body = [
'HttpStatusCode' => 400,
'HttpStatusDescription' => 'Bad Request',
'Message' => 'One or more shipments are invalid',
'Errors' => [
'Message' => 'The first line of the address must be 35 characters or less.',
'Cause' => 'DestinationAddressLine1',
'ErrorCode' => 'E1433', //Invalid field
'ErrorId' => '932478fg83r7gf'
]
];
return new MockResponse([json_encode($body)], ['http_code' => 400]);
}

并像这样使用：

$responses = [CourierServiceTest::mockLoginResponse(), CourierServiceTest::mockShipmentResponseFail()];
$mockClient = new MockHttpClient($responses);
$someService = new SomeService($mockClient, $request->server);
$apiResponse = $someService->orders($tmpOrder1->getId());
$responseArray = json_decode($apiResponse->getContent(), true);
$this->assertEquals(400, $apiResponse->getStatusCode());
$this->assertJson($apiResponse->getContent());

不幸的是，我收到以下错误：

Symfony\Component\HttpClient\Exception\ClientException:HTTP 400返回"；https://api.someservice.net/api/"。

try {
$apiResponse = $this->client->request('POST', $url, [
'headers' => [
'accept' => 'application/json',
'content-type' => 'application/json',
other headers
],
'body' => [
$body
]
]);
} catch (ClientException $exception) {
return $exception->getResponse();
}

Answer 1

以下是我所做的：

$body = '{"data": 1}';
// create the mock response
$client = new MockHttpClient([new MockResponse(
$body,
[
'response_headers' => ['content-type' => 'application/json'],
'http_code' => 400
]
)]);
// do the request
$response = $client->request('GET', $url);
// assert the content 
$this->assertEquals(
$body,
$response->toArray(false)
);

Answer 2

所以你正在创建一个返回错误400的函数，然后当你调用该函数时，你总是会得到一个HTTP 400错误，对吗？

我不明白你想要实现什么，你能澄清一下吗？

Answer 3

我编辑了我的问题，但我意识到getContent((和toArray((抛出ClientException，request((没有，所以无法捕获