Jest SpyOn - 调用实际方法，而不是模拟实现

Question

我到处搜索，试图让我的 Jest 测试正常工作，但我的搜索一无所获。

我有一个 function ( createRoom )，我想对其进行单元测试，并且我想模拟一个方法 ( gameFactory.createNewGameWithInitialPlayer() ) 调用。

游戏服务.ts

const GameService = (games: {[index: string]: Game}) => {
  const playerFactory = PlayerFactory()
  const gameFactory = new GameFactory()

  const createRoom = ({name, device, socketID}: {name: string, device: string, socketID: string}): RoomResponse => {
    const player = playerFactory.createNewPlayer(name, device, socketID)

    if (player && player.id) {
        const game: Game|undefined = gameFactory.createNewGameWithInitialPlayer(player)
        ...
    }
    ...
}

游戏工厂.ts

export class GameFactory {
    createNewGameWithInitialPlayer = (player: Player): Game|undefined => {
        const game = new Game()
        game.spectators[player.id as any as number] = player

        return game
    }
}

游戏服务.test.ts

import * as gameFactory from '../Factories/GameFactory'

describe('Testing Game Service', () => {
    test('createRoom', () => {
        jest.spyOn(gameFactory, 'GameFactory').mockReturnValue({ createNewGameWithInitialPlayer: jest.fn().mockReturnValue(undefined)})

        const response: RoomResponse = gameService.createRoom({
            name: 'Player 1',
            device: DevicesEnum.ios,
            socketID: 'some-socket-id'
        })
    ...
    }
    ...
}

在GameService.test.ts我是 mocking createNewGameWithInitialPlayer方法调用的返回值。 但是，当我运行测试时，运行的是它的实际实现，而不是我的模拟版本。 特别是对于此测试，我希望createNewGameWithInitialPlayer方法返回undefined ，但这并没有发生，它似乎是在调用实际的方法实现。

Answer 1

如果您想覆盖createNewGameWithInitialPlayer并返回您想要的内容，则必须在测试中模拟GameFactory class 的导入。

// Here you are creating your mock and saying the default return Game object
const mockCreateNewGameWithInitialPlayer = jest.fn().mockImplementation(() => new Game());

// Here you say to jest that any file who wants to import "GameFactory"
// will import this fake class
jest.mock('rootofGameFactory/GameFactory', () => ({
    GameFactory: function(){
      return {
        // and when any file wants to execute this method, will execute my mock
        createNewGameWithInitialPlayer: mockCreateNewGameWithInitialPlayer
      }
    }
}))

describe('Testing Game Service', () => {
    test('createRoom', () => {
        const response: RoomResponse = gameService.createRoom({
            name: 'Player 1',
            device: DevicesEnum.ios,
            socketID: 'some-socket-id'
        })
    ...
    }
    ...
}

如果你想改变你的模拟方法的返回值 object，你必须这样做......

test('createRoom 2', () => {
        //Here you say to jest, just ONCE (for this test) return an instance of Game2
        mockCreateNewGameWithInitialPlayer.mockImplementationOnce(() => new Game2())
        const response: RoomResponse = gameService.createRoom({
            name: 'Player 1',
            device: DevicesEnum.ios,
            socketID: 'some-socket-id'
        })
    ...
    }

Answer 2

开玩笑的文件提到

默认情况下，jest.spyOn 也会调用 spied 方法。 这与大多数其他测试库的行为不同。 如果要覆盖原来的function，可以使用jest.spyOn(object, methodName).mockImplementation(() => customImplementation) 或者jest.replaceProperty(object, methodName, jest.fn(() => customImplementation));

所以你可以做类似的事情

jest.spyOn(gameFactory, 'GameFactory').mockImplementation(() => { return undefined })