在具有异步功能的存根中未调用sinon spy [英] sinon spy is not called in a stub with async function

问题描述

使用 sinon 和酶我要测试以下组件：

Using sinon and enzyme I wanna test the following component:

// Apple.js
class Apple extends Component {

  componentDidMount = () => {

    this.props.start();
    Api.get()
      .then(data => {
        console.log(data); // THIS IS ALWAYS CALLED
        this.props.end();
      });
  }

  render () {
    return (<div></div>);
  }
}

如果我只检查 endApy称为，这总是错误的。但是将其包装在 setTimeout 中将使其通过。为什么总是调用 console.log（）但不调用 props.end ？为什么 setTimeout 可以解决此问题？有更好的方法吗？

If I just check endApy.called, it's always false. But wrapping it in a setTimeout will make it pass. Why console.log() is always called but not the props.end? Why setTimeout fixes it? Is there a better way of doing this?

// Apple.test.js
import sinon from 'sinon';
import { mount } from 'enzyme';
import Api from './Api';
import Apple from './Apple';


test('should call "end" if Api.get is successfull', t => {
  t.plan(2);
    sinon
        .stub(Api, 'get')
        .returns(Promise.resolve());

    const startSpy = sinon.spy();
    const endApy = sinon.spy();

    mount(<Apple start={ startSpy } end={ endApy } />);

    t.equal(startSpy.called, true);                    // ALWAYS PASSES 
    t.equal(endSpy.called, true);                      // ALWAYS FAILS
    setTimeout(() => t.equal(endApy.called, true));    // ALWAYS PASSES

    Api.get.restore();
});

推荐答案

Api.get 是异步函数，它返回一个Promise，因此要在测试中模拟异步调用，您需要调用 resolves 函数而不是 returns ：

Api.get is async function and it returns a promise, so to emulate async call in test you need to call resolves function not returns:

使存根返回一个Promise，该Promise会解析为所提供的值。构造Promise时，sinon使用Promise.resolve方法。您负责在不提供Promise的环境中提供polyfill。

Causes the stub to return a Promise which resolves to the provided value. When constructing the Promise, sinon uses the Promise.resolve method. You are responsible for providing a polyfill in environments which do not provide Promise.

sinon
  .stub(Api, 'get')
  .resolves('ok');

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