Question

我试图在我的法典中利用这一点:

const navigation = useNavigation();
navigation.replace( AllFriends )

但我仍犯了一个错误:

Property  replace  does not exist on type  NavigationProp<Record<string, object | undefined>, string, NavigationState, {}, {}> .

我也试图利用这一方法。

const navigation =  useNavigation<StackNavigationProp<{route: {} }>>()
navigation.replace( route );

如下文所示:https://github.com/react-navigation/react-navigation/issues/82<56/a>。

but this gives me an error on replace that expected 2 arguments but got only 1...

I am using "@react-navigation/native": "^5.5.1",

Answer 1

The replace method is only available to screens within StackNavigators. We need a way to explicitly type that the screen is such.

当你在StackNavigator时,你有能力通过界定你的那种方式。路线及其预期得到的参数:

export type StackParams = {
   MyRoute: undefined;
   AnotherRoute: {
      id: number
   }
}

createStackNavigator<StackParams>();

Then in the screen you pass that param list to StackNavigationProp

import { StackNavigationProp } from  @react-navigation/stack/ ;
import { StackParams} form  your/routes ;

const navigation =  useNavigation<StackNavigationProp<StackParams>>()
navigation.replace( route );

以此申明,这是一种圣卡片检查,并提供了你可以通航的路线清单:

replace( AnotherRoute ) //success!
replace( FakeRoute ) //error, this was not defined in our RouteParams

Answer 2

const navigation =  useNavigation<StackNavigationProp<{route: {} }>>()
navigation.replace( route );

Answer 3

如下文所述::https://github.com/react-navigation/react-navigation/issues/7820#issuecomment-6896468

最简单的办法是:

const navigation = useNavigation<StackNavigationProp<ParamListBase>>();

Answer 4

日复一日,仍在使用班级部件,并设有标准的纳格托帕拉姆派

export type NavigatorParamList = {
 ...
}

export type ScreenProps = NavigationProp<NavigatorParamList>

export type ScreenComponentProps<T extends keyof NavigatorParamList> = {
  navigation: NativeStackNavigationProp<NavigatorParamList, T> // changed from navigation: NavigationProp<NavigatorParamList, T>
  route: RouteProp<NavigatorParamList, T>
}

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