React native with typescript-如何使用@React navigation/native with typescript中的useRoute
我正试图从React native with typescript-如何使用@React navigation/native with typescript中的useRoute,typescript,react-native,react-navigation,Typescript,React Native,React Navigation,我正试图从route.params获取我的incident对象,但我不知道如何使typescript识别此道具 以下是导航到myDetail页面并将incident传递到参数的功能: const navigateToDetail = (incident: IncidentProps): void => { navigation.navigate('Detail', { incident }); }; 下面是详细页面代码的一部分,我试图从route.params获取此对象: t
route.params
获取我的incident对象,但我不知道如何使typescript识别此道具
以下是导航到myDetail页面并将incident传递到参数的功能:
const navigateToDetail = (incident: IncidentProps): void => {
navigation.navigate('Detail', { incident });
};
下面是详细页面代码的一部分,我试图从route.params获取此对象:
type IncidentRouteParams = {
incident: IncidentProps;
}
const Detail: React.FC = () => {
const navigation = useNavigation();
const route = useRoute();
const incident = route.params.incident;
我想我需要通过某种方式将这个意外路线图输入传递到const route=useRoute()
提前谢谢
以下是出现错误的图像:
编辑:
我确实喜欢这样做,但我不知道这样做是否正确:
const route = useRoute<RouteProp<Record<string, IncidentRouteParams>, string>>();
const incident = route.params.incident;
constroute=useRoute();
const事件=route.params.incident;
昨天刚做了这件事
TLDR:
首先,您需要使用每个屏幕名称及其接收的参数定义一个类型:
type ParamList = {
Detail: {
incident: IncidentProps;
};
};
然后在RouteProp
中使用该参数和屏幕名称:
const route = useRoute<RouteProp<ParamList, 'Detail'>>();
constroute=useRoute();
以下是解释所有这些的文档您还可以根据所需的ParamList创建一个类型,因此您只需将此类型导入组件并将RouteName作为参数传递
import { RouteProp } from '@react-navigation/native';
export type RootStackParamList = {
Home: undefined;
Feed: { sort: 'latest' | 'top' };
};
export type RootRouteProps<RouteName extends keyof RootStackParamList> = RouteProp<
RootStackParamList,
RouteName
>;
从'@react-navigation/native'导入{RouteProp};
导出类型RootStackParamList={
家:未定义;
提要:{排序:'最新'|'顶部'};
};
导出类型rootRouteProp=RouteProp<
RootStackParamList,
罗特奈
>;
用法:
export const Feed = () => {
const route = useRoute<RootRouteProps<'Feed'>>();
return <Text>{route.params.sort}</Text>
}
export const Feed=()=>{
const route=useRoute();
返回{route.params.sort}
}
这太复杂了。在TS中设置initialParams有一种更简单的方法
设立:
type StackParamList = {
TheSource: { endpoint: string; type: string;};
BenefitDetail: React.FC;
ItineraryMain: {reFetch: boolean;};
Today: React.FC;
};
const Stack = createStackNavigator<StackParamList>();
类型StackParamList={
源:{端点:字符串;类型:字符串;};
BenefitDetail:React.FC;
巡回主站:{reFetch:boolean;};
今天:React.FC;
};
const Stack=createStackNavigator();
然后在使用时:
const TheSourceRoute = (): React.ReactElement => {
return (<StackNavigator id="TheSource" element={
<Stack.Screen name="TheSource" component={WebviewModal} initialParams={{ endpoint: sourceUrl, type: 'directsource' }}
options={{
title: 'The Source',
}}/>
}/>
);
};
const TheSourceRoute=():React.ReactElement=>{
返回(
);
};