如何测试酶的氧化还原成分?

问题描述 投票:-1回答:1

我必须做一些简单的React Enzyme测试。我想检查组件是否已渲染。

import React from 'react';
import { shallow } from 'enzyme';
import ConnSearch from './ConnSearch';

it('renders without errors', () => {
    const component = shallow(<ConnSearch />);
    console.log(component.debug());
});

我有结果:在“ Connect(ConnSearch)”的上下文中找不到“商店”。将根组件包装在中,或者将自定义React上下文提供程序传递给在连接选项中找到相应的React上下文使用者以使用Connect(ConnSearch)。

我的ConnSearch组件:

import React, { Component } from 'react';
import {fetchRoadDetails, fetchUserPoints} from "../../actions";
import {connect} from "react-redux";
import RoadTable from "../../components/RoadTable/RoadTable";
import RoadForm from "../../components/RoadTable/RoadForm";
import style from './ConnSearch.module.scss'
import {getPoints} from "../../reducers";

class ConnSearch extends Component {
    constructor(props){
        super(props);
        this.state = {

        };
    }
    componentDidMount() {
        this.props.fetchUserPoints(this.props.userLogin);
    }


    render() {

        return (
            <div className={style.wrapper}>
                <RoadForm />
                <div className={style.tableWrapper} >
                <RoadTable/>
                </div>
            </div>
        );
    }
}

const mapDispatchToProps=dispatch=>({
    fetchRoadDetails:()=>dispatch(fetchRoadDetails()),
    fetchUserPoints:(user)=>dispatch(fetchUserPoints(user)),
});

const mapStateToProps = (state) => {
    return {
        roads: state.road,
        points:getPoints(state),
        userLogin: state.userLogin,
    };
};


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

我该如何进行测试?我从未做过。

reactjs redux enzyme
1个回答
0
投票

不幸的是,当我将其包装在提供程序中时:

it('renders without errors', () => {
    const component = shallow( <Provider store={store}><ConnSearch/></Provider>);
    console.log(component.debug());

});

我知道了:

  console.log src/views/ConnectionSearch/ConnSearch.test.js:11
    <ContextProvider value={{...}}>
      <Connect(ConnSearch) />
    </ContextProvider>

我想要ConnSearch呈现结构。

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