是什么导致我的redux操作无法正常工作?

问题描述 投票:0回答:1

我是React的初学者,请多多包涵。我一直在努力尝试解决这个问题,现在已经花了几个小时了,只是不明白我在做什么错。

一切正常,只要找到并加载0错误。但是,当我尝试启动一个动作时,绝对没有任何反应。我正在记录状态,并且在应用程序的初始启动之后就没有多余的日志。

所以我要构建的是超级简单的练习应用程序。屏幕上只有一个单词,每个字母本身就是一个组成部分。用户可以单击两个按钮中的任意一个,在单词的开头或结尾添加字母。而已。我仅有的2个动作是添加这些字母。

这是我的主要App.js:

import React from 'react';
import Header from './Header';
import Adder from './Adder';
import CurrentWord from './CurrentWord';
import { connect } from 'react-redux';
import { addLetterStart, addLetterEnd } from '../actions';

class App extends React.Component {

    render() {
        console.log(this.props.word);
        return (
            <div>
                <Header />
                <div style={{ margin: "100px", display: "flex", justifyContent: "center"}}>
                    <Adder onClick={()=> this.props.addLetterStart()} />
                    <CurrentWord />
                    <Adder onClick={()=> this.props.addLetterEnd()} />
                </div>
            </div>
        )
    }
};

const mapStateToProps = (state) => {
    return {
        word: state.word
    }
};

export default connect(mapStateToProps, {addLetterStart, addLetterEnd})(App);

这是我的CurrentWord组件:

import React from 'react';
import SingleLetter from './SingleLetter'
import { connect } from 'react-redux';


class CurrentWord extends React.Component {
    renderLetters() {
        return this.props.word.map(letter => {
            return <SingleLetter key={Math.floor(Math.random() * 100)} letter={letter} />
        });

    }

    render() {
        return (
            <div style={{ display: "flex" }}>
                {this.renderLetters()}
            </div>
        )
    }
};

const mapStateToProps = (state) => {
    return { word: state.word }
};

export default connect(mapStateToProps)(CurrentWord);

这些是我的动作(actions / index.js)

import {ADD_END, ADD_START} from './types';


export const addLetterStart = () => {
    return {
        type: ADD_START
    };
};

export const addLetterEnd = () => {
    return {
        type: ADD_END
    };
};

我的来信者:

import { ADD_END, ADD_START } from '../actions/types';

export default (state = ['X', 'X', 'X', 'X'], action) => {
    switch (action.type) {
        case ADD_START:
        return ['X', ...state];

        case ADD_END:
        return [...state, 'X'];

        default:
            return state;
    }
};

我的reducers / index.js:

import { combineReducers } from 'redux';
import LettersReducer from './LettersReducer';

export default combineReducers({
    word: LettersReducer
});

还有我的主要index.js:

import React from 'react';
import ReactDOM from 'react-dom';
import { Provider } from 'react-redux';
import { createStore } from 'redux';

import App from './components/App';
import reducers from './reducers';

const store = createStore(reducers);

ReactDOM.render(
    <Provider store={store}>
        <App />
    </Provider>,
    document.querySelector('#root')
);

这是我的加法器组件:

import React from 'react';

const Adder = () => {

    return (
        <div className="ui card" style={{
            height: "150px",
            width: "150px",
            display: "flex",
            justifyContent: "center",
            alignItems: "center",
            cursor:"pointer",
            margin: "10px"
        }}>
            <p style={{
                fontSize: "80px", 
                color: "#ececec" }}>+</p>
        </div>
    )

};

export default Adder;

我认为这应该与所有相关。非常感谢您的帮助,谢谢!

javascript reactjs redux store redux-actions
1个回答
2
投票

问题是您要在组件onClick上添加Adder,但Adder不是DOM元素。您需要将该道具转发到可点击的对象

从'react'导入React;

const Adder = ({ onClick }) => {

    return (
        <div className="ui card" style={{
            height: "150px",
            width: "150px",
            display: "flex",
            justifyContent: "center",
            alignItems: "center",
            cursor:"pointer",
            margin: "10px"
        }}>
            <a onClick={onClick}><p c={style={{
                fontSize: "80px", 
                color: "#ececec" }}>+</p></a>
        </div>
    )

};

export default Adder;

如果我的记忆没有失败,则无法单击p标记,因此您可能想用a锚点或button包裹它。否则,您可以直接在p

上执行此操作
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