正如标题所示,我的 React 应用程序遇到问题,我可以点击 Redux Action,但之后就不会点击Reducer。我查看了我过去从事的一个项目,以及这里的几篇文章,但我不确定我的代码有什么问题,导致操作无法命中减速器。我已粘贴下面的代码,但如果我可以提供其他内容,请告诉我。
索引.js:
import React from 'react';
import ReactDom from 'react-dom';
import App from './components/App.jsx';
import { Provider } from 'react-redux';
import { createStore } from 'redux';
import reducer from './reducers/usersRedcuers';
import './index.css';
const store = createStore(reducer);
ReactDom.render(
<Provider store={store}>
<App />
</Provider>, document.getElementById('root')
)
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App.jsx 组件:
import React, { Component } from 'react';
import { connect } from 'react-redux';
import { bindActionCreators } from 'redux';
import Auth from '../modules/Auth';
import { login } from '../actions/usersActions';
class App extends Component {
constructor(props) {
super(props);
this.state = {
auth: null,
token: '',
password: '',
username: '',
}
const {
login,
} = this.props;
}
loginHandler() {
const { password, username } = this.state;
const auth = Auth.isUserAuthenticated()
const token = null;
login(auth, token, password, username);
};
render() {
return (
<div className="App">
<div className="title">
Recipe Box
</div>
<div className="form-inline login-form">
<div className="form-group">
<input
className="form-control"
onChange={e => this.setState({ username: e.target.value })}
placeholder="Username"
/>
<input
className="form-control"
onChange={e => this.setState({ password: e.target.value })}
placeholder="Password"
type="password"
/>
</div>
<button
className="btn btn-success"
onClick={() => this.loginHandler()}
type="button"
>
Login
</button>
</div>
</div>
);
}
}
function mapDispatchToProps(dispatch) {
console.log('dispatching:', dispatch)
return bindActionCreators({login}, dispatch);
}
function mapStateToProps(state) {
return { state }
}
export default connect(mapStateToProps, mapDispatchToProps)(App);
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动作常数:
// AUTH
export const LOGIN = 'LOGIN';
export const LOGOUT = 'LOGOUT';
// USERS
export const ADD_USER = 'ADD_USER';
export const DELETE_USER = 'DELETE_USER';
export const UPDATE_USER = 'UPDATE_USER';
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行动:
import {
ADD_USER,
DELETE_USER,
LOGIN,
LOGOUT,
UPDATE_USER,
} from '../constants/constants';
export const login = (auth, token, password, username) => {
const action = {
type: LOGIN,
auth,
token,
password,
username,
}
console.log('login action:', action)
return action;
}
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减速器:
import {
LOGIN,
LOGOUT,
} from '../constants/constants';
const login = (action) => {
console.log('hitting B')
const { auth, token, password, username } = action;
return {
auth: auth,
token: token,
password: password,
username: username,
}
}
const authControl = (state = [], action) => {
console.log('hitting C: ', action)
let authControl = null;
switch(action.type) {
case LOGIN:
authControl = [...state, login(action)]
console.log('authControl:'. authControl);
return authControl;
default:
console.log('hittibbng default', state)
return state;
}
}
export default authControl;
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小智 5
在 App.jsx 组件中,您应该使用作为 prop 传递给组件的操作,而不是直接调用该操作。
登录处理程序应该如下所示:
loginHandler() {
const { password, username } = this.state;
const auth = Auth.isUserAuthenticated()
const token = null;
this.props.login(auth, token, password, username);
};
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