在异步函数中使用 ts-mockito 引发错误答案

【问题标题】：throw an error using ts-mockito in async function在异步函数中使用 ts-mockito 引发错误
【发布时间】：2020-04-02 01:59:39
【问题描述】：

我正在尝试使用 ts-mockito 测试 API 端点。事情进展顺利，直到我开始异步...... 我的测试：

it('should throw an error if the address is not valid',()=>{
  const mockedGeolocationService: GoogleGeolocationService = mock(GoogleGeolocationService);
  when(mockedGeolocationService.getLocation(address)).thenThrow(new Error("the address provided is not valid"));
  const geolocationService: IGeolocationService = instance(mockedGeolocationService);

  const addressService: AddressService = new AddressService(geolocationService, new MongoAddressRepository());
  expect(() => addressService.storeAddress(address)).to.throw("the address provided is not valid");
});

服务：

public storeAddress = (address: Address): void => {
  const location = this.geolocationService.getLocation(address);
  address.setLocation(location);
  this.addressRepository.store(address);
}

到目前为止，一切正常。但是当我开始实现地理定位服务时，我不得不将它声明为一个 Promise，因为它执行的是一个 http 请求。

public storeAddress = async (address: Address): Promise<void> => {
  const location = await this.geolocationService.getLocation(address);
  address.setLocation(location);
  this.addressRepository.store(address);
}

然后我无法再捕获抛出的错误，如果它毕竟被抛出...... 任何线索我应该如何捕获或抛出这个错误？提前致谢。

【问题讨论】：

标签： javascript typescript mocha.js chai ts-mockito

【解决方案1】：

我想出了如何处理错误，我留下答案以防万一它对某人有所帮助。问题是因为它是一个异步函数，所以它不会捕获错误。它必须手动处理并比较错误（作为字符串），如下所示：

it('should throw an error if the address is not valid',(done)=>{
    const mockedGeolocationService: GoogleGeolocationService = mock(GoogleGeolocationService);
    when(mockedGeolocationService.getLocation(address)).thenThrow(new Error("the address provided is not valid"));
    const geolocationService: IGeolocationService = instance(mockedGeolocationService);

    const addressService: AddressService = new AddressService(geolocationService, new MongoAddressRepository());
    addressService.storeAddress(address).catch(error => {
        expect(error.toString()).to.be.equal(new Error("the address provided is not valid").toString());
        done();
    });
});

【讨论】：