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创建React应用程序,由于1个错误,所有测试用例都失败

jestjs enzyme react-redux reactjs javascript

Atul kumar singh · 技术社区 · 6 年前

我收到此错误-:

不变冲突:在“连接(登录)”的上下文或道具中找不到“存储”。将根组件包装在中,或者将“store”作为prop显式传递给“Connect(Login)”

我的索引。js公司

import React from 'react';
import ReactDOM from 'react-dom';
import './fonts/stylesheet.css';
import './index.css';
import Routes from "./routes";
import registerServiceWorker from './registerServiceWorker';
import reducer from './reducers';
import { createStore, applyMiddleware } from 'redux';
import { Provider } from 'react-redux';
import thunk from 'redux-thunk';

const middleware = [ thunk ]
const store = createStore(
    reducer,
    applyMiddleware(...middleware)
)

ReactDOM.render(
    <Provider store={store}>
        <Routes />
    </Provider>,
    document.getElementById('root')
);
registerServiceWorker();

和登录。js是

import React, { Component } from 'react';
import { connect } from 'react-redux';
import { setLoginStatusAndPartnerName } from '../actions';
import Header from './header';
import Footer from './footer';

class Login extends Component {

  constructor() {
    super();
    this.state = {
      emailId: "",
      isError: false,
      partner: "",
      loginStatus: false
    };
    this.checkEmail = this.checkEmail.bind(this);
    this.handleChange = this.handleChange.bind(this);
  }


  handleChange({ target }) {
    this.setState({
      [target.name]: target.value,
      isError: false
    });
  }

  render() {
    return (
      <div>
      <div className="App">
        <Header/>
          <div className="login-container">
          <form onSubmit={this.checkEmail}>
             <div className="email-id">
               <label>
                 Email Address
               </label>
               <input type="text" value={ this.state.emailId }
               onChange={ this.handleChange } name="emailId"/>
             </div>
             </form>
           </div>
       </div>
        <Footer/>
        </div>
    );
  }
}

function mapStateToProps(state) {
  const { loginStatus= false, partner='' } = state;
  return{
    loginStatus,
    partner
  }
}

function mapDispatchToProps(dispatch){
  return {
    sendLoginStatusAndInfo: (loginStatus , partner) => {
      dispatch(setLoginStatusAndPartnerName(loginStatus , partner));
    }
  }
}

export default connect(mapStateToProps, mapDispatchToProps)(Login);

我的登录名。测验js公司

import React from 'react';
import ReactDOM from 'react-dom';
import { shallow, mount, render } from 'enzyme';
import Login from '../components/login';

describe('Login Component render', () => {

  it('login renders without crashing', () => {
    const div = document.createElement('div');
    ReactDOM.render(<Login />, div);
  });

  it('checking whether the login form is present or not', () => {
   expect(shallow(<Login />).exists(<form></form>)).toBe(true)
  })

  it('renders a label of login form', () => {
   expect(shallow(<Login />).find('label').length).toEqual(1)
  })
  it('renders input box of login', () => {
   expect(shallow(<Login />).find('input').length).toEqual(1)
  })

  it('renders the login information text', () => {
  expect(shallow(<Login />).find('.login-info-text').length).toEqual(1)
  })
});

我不确定我做错了什么,但所有的测试用例都以相同的错误失败。

2 回复 | 直到 6 年前

Andrew Miroshnichenko 6 年前

这是因为你试图测试 connect 教育部 Login 组件,但您可能应该测试原始组件。添加 export 组件定义,如下所示:

export class Login extends Component

和导入非默认值( 连接 ed)但未经加工的组件:

import {Login} from '../components/login';

这里是 docs to read

Achraf C. 6 年前

import { HashRouter, Route, Switch } from 'react-router-dom';

ReactDOM.render((
  <Provider store={store}>
    <HashRouter history={history}>
      <Switch>
        <Route path="/" name="Home" component={Full} />
      </Switch>
    </HashRouter>
  </Provider>
), document.getElementById('root'));

这是CoreUi模板的最佳实践。您重新投影的管线是否可能不返回组件?