[英]How extract state change action to a named function with typed parameters? NgRx, Angular
有没有办法创建一个命名状态更改函数,具有适当的参数类型,在创建减速器时将在on
方法中接受?
我想创建onLoginSuccessful
函数,该函数将处理状态更改并可以传递给减速器中的on
方法。
但是当我尝试创建onLoginSuccessful
时,出现 TS 编译错误。
//== actions.ts file ==//
export const loginSuccessful = createAction(
'[Login page] Login successful',
props<{authToken: string}>()
);
//== reducer.ts file ==//
export const initialState: AuthState = {
authToken: null
};
// this works
export const reducer = createReducer(
initialState,
on(loginSuccessful, (state, action) => {
return {
...state,
authToken: action.authToken
};
})
);
// this does NOT work
// creating named function onLoginSuccess with typed params
function onLoginSuccess(state: AuthState, action: typeof loginSuccessful): AuthState {
return {
...state,
authToken: action.authToken
};
}
export const reducer = createReducer(
initialState,
on(loginSuccessful, onLoginSuccess) // <-- here on "onLoginSuccess" param throws TS compiler an error
);
TS编译错误:
'(state: AuthState, action: ActionCreator<"[登录页面] 登录成功", (props: { authToken: string; }) => { authToken: string; } & TypedAction<"[登录页面] 登录成功的参数">>) => AuthState' 不能分配给类型为 'OnReducer<AuthState, [ActionCreator<"[Login page] 登录成功的参数", (props: { authToken: string; }) => { authToken: string; } & TypedAction<"[登录页面]登录成功">>]>'. 参数“action”和“action”的类型不兼容。 输入 '{ authToken: 字符串; } & TypedAction<"[登录页面] 登录成功"> & { type: "[登录页面] 登录成功"; }' 不可分配到类型 'ActionCreator<"[登录页面] 登录成功", (props: { authToken: string; }) => { authToken: string; } & TypedAction<"[登录页面]登录成功">>'. 输入 '{ authToken: 字符串; } & TypedAction<"[登录页面] 登录成功"> & { type: "[登录页面] 登录成功"; }' 不可分配给类型 '(props: { authToken: string; }) => { authToken: string; } & TypedAction<"[登录页面]登录成功">'。 输入 '{ authToken: 字符串; } & TypedAction<"[登录页面] 登录成功"> & { type: "[登录页面] 登录成功"; }' 不匹配签名 '(props: { authToken: string; }): { authToken: string; } & TypedAction<"[登录页面]登录成功">'.ts(2345)
尝试使用ActionType<typeof loginSuccessful>
:
import { ActionType } from "@ngrx/store";
// ...
function onLoginSuccess(state: AuthState, action: ActionType<typeof loginSuccessful>): AuthState {
return {
...state,
authToken: action.authToken
};
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.