这个减速机好吗：
function someReducer(state = initialState, action) {
if (action.type === SOME_ACTION) {
const newState = Object.assign( {}, state );
// ...
// doing whatever I want with newState
// ...
return newState;
}
return state;
}
如果没问题，为什么我们需要所有这些不可变的库来使我们的生活复杂化。
ps 只是试图理解 Redux 和不变性
export default function (state = initialState, action) {
const actions = {
SOME_ACTION: () => {
return {
...state
}
},
ANOTHER_ACTION: () => {
return {
...state
error: action.error
}
},
DEFAULT: () => state;
}
return actions[action.type] ? actions[action.type]() : actions.DEFAULT();
}我更喜欢这样做。我不是 switch 语句的忠实粉丝。
标准方法是在减速器中使用switch/casewith spread 语法 ( ...)。
export default function (state = initialState, action) {
switch (action.type) {
case constants.SOME_ACTION:
return {
...state,
newProperty: action.newProperty
};
case constants.ERROR_ACTION:
return {
...state,
error: action.error
};
case constants.MORE_DEEP_ACTION:
return {
...state,
users: {
...state.users,
user1: action.users.user1
}
};
default:
return {
...state
}
}
}
然后，您可以使用 ES6 扩展语法返回您的旧状态以及您想要更改/添加的任何新属性。
您可以在此处阅读有关此方法的更多信息... https://redux.js.org/recipes/using-object-spread-operator
我发现了一些我非常喜欢的东西：
import createReducer from 'redux-starter-kit';
const someReducer = createReducer( initialState, {
SOME_ACTION: (state) => { /* doing whatever I want with this local State */ },
SOME_ANOTHER_ACTION: (state) => { /* doing whatever I want with local State */ },
THIRD_ACTION: (state, action) => { ... },
});
如果您的状态具有嵌套的对象或数组，Object.assign或者...将复制对旧状态变量的引用，则可能会导致一些问题。这就是为什么一些开发人员使用不可变库的原因，因为在大多数情况下状态具有深度嵌套的数组或对象。
function someReducer(state = initialState, action) {
if (action.type === SOME_ACTION) {
const newState = Object.assign( {}, state );
// newState can still have references to your older state values if they are array or orobjects
return newState;
}
return state;
}