35

目前我正在使用带有反应钩子的功能组件。但我无法useState完全测试钩子。考虑这样一个场景,在useEffect钩子中我正在做一个 API 调用并在useState. 对于 jest/enzyme,我已经模拟了要测试的数据,但我无法useState在 jest 中设置初始状态值。

const [state, setState] = useState([]);

我想将初始状态设置为开玩笑的对象数组。我找不到任何类似于类组件的 setState 函数。

4

6 回答 6

24

您可以模拟React.useState以在测试中返回不同的初始状态:

// Cache original functionality
const realUseState = React.useState

// Stub the initial state
const stubInitialState = ['stub data']

// Mock useState before rendering your component
jest
  .spyOn(React, 'useState')
  .mockImplementationOnce(() => realUseState(stubInitialState))

参考: https ://dev.to/theactualgivens/testing-react-hook-state-changes-2oga

于 2019-08-06T21:00:48.403 回答
18

首先,您不能在组件中使用解构。例如,您不能使用:

import React, { useState } from 'react';
const [myState, setMyState] = useState();

相反,您必须使用:

import React from 'react'
const [myState, setMyState] = React.useState();

然后在你的test.js文件中:

test('useState mock', () => {
   const myInitialState = 'My Initial State'

   React.useState = jest.fn().mockReturnValue([myInitialState, {}])
   
   const wrapper = shallow(<MyComponent />)

   // initial state is set and you can now test your component 
}

如果您在组件中多次使用 useState 挂钩:

// in MyComponent.js

import React from 'react'

const [myFirstState, setMyFirstState] = React.useState();
const [mySecondState, setMySecondState] = React.useState();

// in MyComponent.test.js

test('useState mock', () => {
   const initialStateForFirstUseStateCall = 'My First Initial State'
   const initialStateForSecondUseStateCall = 'My Second Initial State'

   React.useState = jest.fn()
     .mockReturnValueOnce([initialStateForFirstUseStateCall, {}])
     .mockReturnValueOnce([initialStateForSecondUseStateCall, {}])
   
   const wrapper = shallow(<MyComponent />)

   // initial states are set and you can now test your component 
}
// actually testing of many `useEffect` calls sequentially as shown
// above makes your test fragile. I would recommend to use 
// `useReducer` instead.
于 2020-12-17T04:45:44.420 回答
10

如果我没记错的话,你应该尽量避免模拟内置的钩子,比如useStateand useEffect。如果使用酶很难触发状态变化invoke(),那么这可能表明您的组件将从被分解中受益。

于 2019-08-06T21:07:20.323 回答
5
//Component    
const MyComponent = ({ someColl, someId }) => {
     const [myState, setMyState] = useState(null);

     useEffect(() => {loop every time group is set
         if (groupId) {
             const runEffect = async () => {
                  const data = someColl.find(s => s.id = someId);
                  setMyState(data);
             };
             runEffect();
         }
     }, [someId, someColl]);

     return (<div>{myState.name}</div>);
};

// Test
// Mock
const mockSetState = jest.fn();
jest.mock('react', () => ({
    ...jest.requireActual('react'),
    useState: initial => [initial, mockSetState]
}));
const coll = [{id: 1, name:'Test'}, {id: 2, name:'Test2'}];

it('renders correctly with groupId', () => {
    const wrapper = shallow(
        <MyComponent comeId={1} someColl={coll} />
    );
    setTimeout(() => {
        expect(wrapper).toMatchSnapshot();
        expect(mockSetState).toHaveBeenCalledWith({ id: 1, name: 'Test' });
    }, 100);
});
于 2020-10-01T16:28:37.953 回答
5
  • 下面的函数将返回状态
const setHookState = (newState) => jest.fn().mockImplementation(() => [
  newState,
  () => {},
]);
  • 在下面添加以使用反应

const reactMock = require('react');

在你的代码中,你必须习惯React.useState()这个工作,否则它不会工作

const [arrayValues, setArrayValues] = React.useState();

const [isFetching, setFetching] = React.useState();

  • 然后在您的测试中添加以下模拟状态值

reactMock.useState = setHookState({ arrayValues: [], isFetching: false, });

灵感:后藤

于 2020-02-27T15:55:36.547 回答
1

我花了很多时间,但找到了在我的应用程序中测试多个 useState 的好解决方案。

export const setHookTestState = (newState: any) => {
  const setStateMockFn = () => {};
  return Object.keys(newState).reduce((acc, val) => {
    acc = acc?.mockImplementationOnce(() => [newState[val], setStateMockFn]);
    return acc;
  }, jest.fn());
};

其中 newState 是我的组件中具有状态字段的对象;

例如:

React.useState = setHookTestState({
      dataFilter: { startDate: '', endDate: '', today: true },
      usersStatisticData: [],
    });
于 2022-01-14T10:23:58.860 回答