Question

如何使用spyOn开玩笑地测试此代码？有人可以为我提供适当的解决方案吗？

.result[0]

我已经尝试过了：

const controller = {};

controller.fetchCars = async (req, res) => {
  const result = await getCarsService(req.body);
  res.json(result);
};

并给我这个错误：因为它不是一个函数，所以无法监视getCarsService属性。未给定

getCarsService（）从/ services导入，看起来像这样：

    it('fetchCars should call getCarsService', () => {
      const spy = jest.spyOn(carsController, 'getCarsService').mockImplementation(() => {});
      const req = {};
      const res = {};

      carsController.fetchCars(req, res);
      expect(spy).toBeCalled()
      spy.mockRestore();
    });

Answer 1

最简单的解决方案是将代码分成每个文件，然后仅使用jest.mock模拟getCarsService方法。

index.js

import getCarsService from './getCarsService';

const controller = {
    fetchCars: async (req, res) => {
        const result = await getCarsService(req.body);
        return res.json(result);
    },
};

export default controller;

getCarsService.js

function getCarsService(body) {
    return new Promise(resolve => resolve(1));
}

export default getCarsService;

index.spec.js

import controller from './index';
import getCarsService from './getCarsService';

jest.mock('./getCarsService', () => jest.fn().mockResolvedValue(1));

it('fetchCars should call getCarsService', async () => {
    const req = {
        hello: 'world',
    };
    const res = {
        json: jest.fn(),
    };

    await controller.fetchCars(req, res);
    expect(getCarsService).toBeCalled();
    expect(res.json).toBeCalledWith(1);
});

如果您对使用间谍一无所知，可以通过监视默认导出来做到这一点。

import controller from './index';
import * as getCarsService from './getCarsService';

it('fetchCars should call getCarsService', async () => {
    const req = {
        hello: 'world',
    };
    const res = {
        json: jest.fn(),
    };

    const spy = jest.spyOn(getCarsService, 'default');
    spy.mockResolvedValue(2);

    await controller.fetchCars(req, res);
    expect(res.json).toBeCalledWith(2);
});

用SpyOn开玩笑

1 个答案: