如何连接reducers逻辑

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，我在多个操作之间共享逻辑(isLogging属性)。但是，当loginSuccesfull操作被分派时，只会触发其中一个reducers。

可以连接共享逻辑，而不是编写：

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

这是不可能的(目前)，只有相同动作类型的最后一个注册动作接收更新。

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，这可能会在下一个版本中“退出”。