如何在使用middy时调用处理程序函数？

Question

我在middy中使用一个http请求函数作为处理函数，然后在启动http请求之前使用ssm中间件获取一些ssm参数。就像这样：

  const makeThirdPartyServiceRequest = middy(async ({ params }) => {
  logger.info(`SENDING Request to ${endpoint} API`)
  const url = `https://someurltoathirdpartyservice`
  const options = {
    method: 'POST',
    body: params
  }

  return helpers.makeRequest(url, options)
})
makeThirdPartyServiceRequest.use(ssm(......))

然而，在我的jest单元测试中，我试图模拟makeThirdPartyServiceRequest并明确表示它应该解析为一个值：

jest.mock('../src/thirdPartyService', () => ({
  __esModule: true,
  default: {
    ...(jest.requireActual('../src/thirdPartyService') as { default: {} }).default,
    makeThirdPartyServiceRequest: jest.fn()
  }
}))
export {}
import thirdPartyService from '../src/thirdPartyService'

然后在测试中我说：

describe('makeThirdPartyServiceRequest()', () => {
  it('should makeThirdPartyServiceRequest', async () => {
    // Given

    // })
    const mockedThirdPartyServiceRequest = mocked(thirdPartyService.makeThirdPartyServiceRequest).mockResolvedValue({})
    // When
    const result = await thirdPartyService.makeThirdPartyServiceRequest(something)
    // Then
    expect(mockedThirdPartyServiceRequest).toHaveBeenCalledTimes(1)
    expect(mockedThirdPartyServiceRequest.mock.calls[0][0].params.toString()).toBe(expectedParams)
  })
})

然而，由于某些原因，middy中间件仍在被调用，我显然不想调用这个中间件，我试图模仿它.我做错什么了？

Answer 1

您需要模拟middy，使它成为一个无用的函数。该函数将函数作为参数并返回该参数。

import thirdPartyService from '../src/thirdPartyService'

jest.mock('@middy/core', () => {
  return (handler) => {
    return {
      use: jest.fn().mockReturnValue(handler), // ...use(ssm()) will return handler function
    }
  }
})

describe('thirdPartyService()', () => {
  beforeEach(() => {
    jest.spyOn(helpers, 'makeRequest') // spy on helpers unit
  })

  describe('makeThirdPartyServiceRequest', () => {
    it('should make a request with correct parameters', async () => {
      // Given
      const url = `https://someurltoathirdpartyservice`
      const params = 'any params'
      const apiResponse = 'any response'
      mocked(helpers.makeRequest).mockResolvedValue(apiResponse)

      // When
      const actual = await thirdPartyService.makeThirdPartyServiceRequest(params)

      // Then
      expect(actual).toBe(apiResponse)
      expect(helpers.makeRequest).toHaveBeenCalledWith(
        url,
        {
          method: 'POST',
          body: params
        }
      )
    })
  })
})

Answer 2

黄德夫的回答也是有效的，但我也会回答我如何继续。

如果你完全想模仿侏儒，你可以像下面这样嘲笑：

    jest.mock('@middy/core', () => {
      return (handler) => {
        return {
          use: jest.fn().mockImplementation(() => {
            // ...use(ssm()) will return handler function
            return {
              before: jest.fn().mockReturnValue(handler)
            }
          })
        }
      }
    })

但是，如果您不想完全模拟middy，您可以从之前调用的middy/util中模拟异步getInternal函数，如下所示：

    jest.doMock('@middy/util', () => ({
      ...(jest.requireActual('@middy/util') as {}),
      getInternal: jest.fn()
    }))
import { getInternal } from '@middy/util'

然后在测试中

    describe('thirdPartyService()', () => {
      beforeEach(() => {
        jest.spyOn(helpers, 'makeRequest') // spy on helpers unit
      })
    
      describe('makeThirdPartyServiceRequest', () => {
        it('should make a request with correct parameters', async () => {
          // Given
          const url = `https://someurltoathirdpartyservice`
          const params = 'any params'
          const apiResponse = 'any response'
          mocked(getInternal).mockResolvedValue({
          twilioSecrets: { accountSid: 'someSID', serviceId: 
          'someServiceID', token: 'someToken' }
          })
          mocked(helpers.makeRequest).mockResolvedValue(apiResponse)
    
          // When
          const actual = await thirdPartyService.makeThirdPartyServiceRequest(params)
    
          // Then
          expect(actual).toBe(apiResponse)
          expect(helpers.makeRequest).toHaveBeenCalledWith(
            url,
            {
              method: 'POST',
              body: params
            }
          )
        })
      })
    })

这将模拟middy的异步部分。