Question

const authReducer = createReducer(
  initialState,
  on(startLogin, (state) => ({...state, isLogging: true})),
  on(loginSuccessful, (state) => ({...state /*do sth other */})),
  on(loginSuccessful, loginFailed, (state) => ({...state, isLogging: false}))
);

在上面的示例中，我在多个操作（isLogging属性）之间共享了逻辑。但是，在分派loginSuccesfull操作时，将仅触发其中的一个reduce。

可以连接共享逻辑而不用编写：

const authReducer = createReducer(
  initialState,
  on(startLogin, (state) => ({...state, isLogging: true})),
  on(loginSuccessful, (state) => ({...state, isLogging: false /*do sth other */})),
  on(loginFailed, (state) => ({...state, isLogging: false}))
);

Answer 1

（目前）这是不可能的，只有相同动作类型的最后一个注册动作会接收更新。

const authReducer = createReducer(
  initialState,
  on(startLogin, (state) => ({...state, isLogging: true})),
  on(loginSuccessful, (state) => ({...state /*do sth other */})),
  on(loginSuccessful, loginFailed, (state) => ({...state, isLogging: false})) // only this one will be triggered for loginSuccessful
);

我们刚刚合并了一个PR，以使您的用例成为可能，下一个版本可能会“淘汰”。

如何连接减速器逻辑

1 个答案: