javascript - removeTodo 操作未从状态中删除

Question

我不确定我哪里出了问题，但是当我触发removeTodo操作时，什么也没有发生。我相信这是我的一个传递错误，但我不太明白。我认为通过将我的操作设置为将 todo 作为其有效负载，然后在我的调度中我向该操作提供 todo.id，它可以删除所述 id。不太明白为什么这行不通。

TodoItem.js

import React, { Component } from 'react';
import { connect } from 'react-redux';
import { removeTodo } from '../actions';
import '../../css/Todo.css';

const mapDispatchToProps = dispatch => {
  return {
      removeTodo: todo => dispatch(removeTodo(todo.id))
    };
  };

const mapStateToProps = state => {
  return {todos: [...state.todos]};
};

class ConnectedTodoItem extends Component {
  render() {
    const {handleToggle, todoId} = this.props;
    const mappedTodos = this.props.todos.map((todo, index) => (
      <div className='todo-item'>
        <span onClick={handleToggle} index={index} id={todoId}>
          {todo.title}
        </span>
        <button type='submit' className='rem-btn' id={todoId} onClick={this.props.removeTodo}>X</button>
      </div>
    ));

    return (
      mappedTodos
    );
  }
}

const TodoItem = connect(mapStateToProps, mapDispatchToProps) (ConnectedTodoItem);

export default TodoItem;

reducer .js

import { ADD_TODO } from '../constants/action-types'; 
import { REMOVE_TODO } from '../constants/action-types';
import uuidv1 from 'uuid';

const initialState = {
  todos: []
};

const rootReducer = (state = initialState, action) => {
  switch (action.type) {
    case ADD_TODO:
    return {
        ...state,
        todos: [...state.todos,
          {
            title: action.payload.inputValue,
            id: uuidv1()
          }]
    }

    case REMOVE_TODO:
    return {
      ...state,
      todos: [...state.todos.filter(todo => todo.id  !== action.payload)]
    }

    default:
      return state;
  }
}

export default rootReducer;

actions.js

import { ADD_TODO } from '../constants/action-types';
import { REMOVE_TODO } from '../constants/action-types';

export const addTodo = (todo) => (
  {
    type: ADD_TODO,
    payload: todo
  }
);

export const removeTodo = (todo) => (
  {
    type: REMOVE_TODO,
    payload: todo.id
  }
)

Answer 1

要添加到Andy的答案中，似乎您还在调度调用中提取了id，但您已经在定义removeTodo的actions.js中执行了此操作，因此除非您在todo.id中有嵌套的id，您需要将其从任一位置删除。