如何解决TypeScript Redux Action 创建函数返回类型
使用 FSA 规则创建动作函数。
未指定返回类型,eslint中显示warning。
是否有一种简单的方法来指定除任何类型之外的类型?
const ADD_Todo = 'todos/ADD_Todo' as const;
const TOGGLE_Todo = 'todos/TOGGLE_Todo' as const;
const REMOVE_Todo = 'todos/REMOVE_Todo' as const;
// Missing return type on function!!!?
export const addTodo = (text: string) => ({
type: ADD_Todo,payload: text,});
// Missing return type on function!!!?
export const toggletodo = (id: number) => ({
type: TOGGLE_Todo,payload: id,});
// Missing return type on function!!!?
export const removetodo = (id: number) => ({
type: REMOVE_Todo,});
type TodosAction = ReturnType<typeof addTodo> | ReturnType<typeof toggletodo> | ReturnType<typeof removetodo>;
export type Todo = {
id: number;
text: string;
done: boolean;
};
export type Todosstate = Todo[];
const initialState: Todosstate = [
{ id: 1,text: 'Hi',done: true },{ id: 2,text: 'Every',{ id: 3,text: 'one',done: false },];
function todos(state: Todosstate = initialState,action: TodosAction): Todosstate {
switch (action.type) {
case ADD_Todo: {
const nextId = Math.max(...state.map((todo) => todo.id)) + 1;
return state.concat({
id: nextId,text: action.payload,done: false,});
}
case TOGGLE_Todo:
return state.map((todo) => (todo.id === action.payload ? { ...todo,done: !todo.done } : todo));
case REMOVE_Todo:
return state.filter((todo) => todo.id !== action.payload);
default:
return state;
}
}
export default todos;
解决方法
你能写一个泛型吗?像这样:
type ActionCreate<TP> = (p: TP) => { type: string,payload: TP };
const addTodo: ActionCreate<string> = (v) => ({
type: 'ADD',payload: v
})
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