如何监视在玩笑中导入的函数

Question

在下面放入一个小片段：

import xyz from '../xyz'
function calculate() {
  return xyz(arg1, arg2).catch((err) => {
    func1()
    func2()
  })
}
export default calculate

我只是想断言 xyz 是开玩笑的。我该怎么做？

我尝试了以下但不起作用：

import * as myModule from '../xyz'
import calculate from '../../calculate'
const mock = jest.spyOn(myModule, 'xyz')
mock.mockReturnValue('mocked value')
const op = calculate()
expect(op).toBe('mocked value')

这给了我以下错误：

Cannot spy the xyz property because it is not a function; undefined given instead

Answer 1

您可以像这样模拟模块：

import calculate from '../../calculate'
jest.mock('../xyz', ()=> () => Promise.resolve('mocked value'))

it('does something', async()=>{
  const op = await calculate()
  expect(op).toBe('mocked value')
})

如果您需要来自 mock 的不同 return 值，您需要 mock 模块以便它 return 成为间谍。然后你必须导入模块，你可以在测试期间设置 return 值：

import calculate from '../../calculate'
import myModule from '../xyz'
jest.mock('../xyz', ()=> jest.fn())

it('does something', async() => {
  myModule.mockImplementation(() => () =>  Promise.resolve('mocked value'))

  const op = calculate()
  expect(op).toBe('mocked value')
})

it('does something else', async() => {
  myModule.mockImplementation(() => () =>  Promise.resolve('another value'))
  const op = await calculate()
  expect(op).toBe('another value')
})


it('does fail', async() => {
  myModule.mockImplementation(() => () =>  Promise.reject('some Error')
  try{
    const op = await calculate()
  }catch (e){
    expect(e).toBe('some Error')
  }
})

如何监视在玩笑中导入的函数

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