为反应导航道具添加强类型

Question

我在我的 react-native 项目 (expo) 中使用 typescript。

该项目使用 react-navigation，所以在我的屏幕上我可以设置 navigationOptions 并且我可以访问 prop navigation。

现在我正在尝试对这些进行强类型化，以便获得有关可以设置哪些属性的提示。

interface NavStateParams {
    someValue: string
}

interface Props extends NavigationScreenProps<NavStateParams> {
   color: string
}

class Screen extends React.Component<Props, any> {
    // This works fine
    static navigationOptions: NavigationStackScreenOptions = {
        title: 'ScreenTitle'
    }
    // Does not work
    static navigationOptions: NavigationStackScreenOptions = ({navigation, screenProps }) => ({
        title: navigation.state.params.someValue
    })
}

将 react-navigation 作为组件的 props 处理的最佳方法是什么。

Answer 1

只需将 NavigationType 添加到您的 Props 中，如下所示：

    import { StackNavigator, NavigationScreenProp } from 'react-navigation';

    export interface HomeScreenProps {
      navigation: NavigationScreenProp<any,any>
    };

    export class HomeScreen extends React.Component<HomeScreenProps, object> {

      render() {
        return (
          <View style={styles.container}>       
            <Button
              title="Go to Details"
              onPress={() => this.props.navigation.navigate('Details')}
            />
          </View>
        );
      }
    }

Answer 2

如果您传递的是由

定义的navigation prop

let navigation = useNavigation()

对于一个组件，最好的输入方式是：

import {NavigationProp, ParamListBase} from '@react-navigation/native';

navigation: NavigationProp<ParamListBase>

更新：

这是一种更好的强导航输入方法，使用最新的@react-navigation 版本 (6.x)

完整示例：

import {NativeStackNavigationProp} from '@react-navigation/native-stack';

type RootStackParamList = {

   Pdp: undefined; //current screen

   PdpComments: {slug: string}; // a screen that we are 
// navigating to, in the current screen,
// that we should pass a prop named `slug` to it

   Sellers: {data: Array<string>};

   Favorites: undefined; // a screen that we are navigating to 
// in the current screen, that we don't pass any props to it
};

interface IPdpPageProps {
   navigation: NativeStackNavigationProp<RootStackParamList, 'Pdp'>;
}

// Since our screen is in the stack, we don't need to 
// use `useNavigation()` to provide the `navigation` to
// our component, we just need to read it as a prop

function Pdp({navigation}: IPdpPageProps) {
   return ...
}

Answer 3

这行得通：

static navigationOptions = ({ navigation }: NavigationScreenProps) => ({
  ...
})

Answer 4

我认为react-navigation 5.X 现在更简单了。以下是如何键入提示 navigation 传递给屏幕/组件的道具：

export default class Header extends React.Component<{
    navigation: StackNavigationHelpers;
}> {
...
}

Ps：用这些版本测试过

"@react-navigation/native": "^5.2.3",
"@react-navigation/stack": "^5.3.1",

Answer 5

我有同样的问题，这是我的解决方案：

import * as React from 'react'
import { NavigationScreenProps, NavigationStackScreenOptions } from 'react-navigation'

interface NavStateParams {
  someValue: string
}

// tslint:disable-next-line:no-any
type NavigationOptionsFn<TParams=any> = (props: NavigationScreenProps<TParams>) => NavigationStackScreenOptions

class Screen extends React.Component {
  // This should works fine
  static navigationOptions: NavigationOptionsFn<NavStateParams> = ({ navigation, screenProps }) => ({
    title: navigation.state.params.someValue
  })
}

您可能希望将NavigationOptionsFn<TParams> 类型声明到某个d.ts 文件中以使其在全局范围内工作。

Answer 6

 yarn add --dev @types/jest @types/react-navigation

import { NavigationScreenProps } from "react-navigation";

export interface ISignInProps extends NavigationScreenProps<{}> { userStore: IUserStore }

export class SignInScreen extends React.Component { .... }

Answer 7

public static navigationOptions: NavigationScreenConfig<NavigationStackScreenOptions> = 
    ({navigation}) => ({/* Your options... */})

Answer 8

如果有人在扩展 NavigationScreenProps 时仍然遇到此问题，那么您可以正确输入 navigationOptions 等以及您自己的道具：

interface Props extends NavigationScreenProps {
  someProp: string;
  anotherProp: string;
}

export const SomeGreatScreen: NavigationScreenComponent<NavigationParams, {}, Props> = ({
  someProp,
  anotherProp,
}) => {
...
};

而NavigationScreenComponent<Props> 导致解构属性{ someProp, anotherProp } 的类型错误，无法识别道具的扩展，NavigationScreenComponent<NavigationParams, {}, Props> 做到了。这似乎是由于需要将扩展​​的 props 类型作为第三个参数发送：

  export type NavigationScreenComponent<
    Params = NavigationParams,
    Options = {},
    Props = {}
  > = React.ComponentType<NavigationScreenProps<Params, Options> & Props> & {
    navigationOptions?: NavigationScreenConfig<Options>;
  };

来自react-navigation.d.ts

Answer 9

您可以在 Props 的界面中直接扩展 NavigationScreenProps，而不是手动描述所有 navigation functions（例如：导航）。

就我而言，必须阻止 eslint 产生错误。

import { StackNavigator, NavigationScreenProps } from 'react-navigation';

export interface HomeScreenProps extends NavigationScreenProps {
/* your custom props here */
};

export class HomeScreen extends React.Component<HomeScreenProps, object> {

  render() {
    return (
      <View style={styles.container}>       
        <Button
          title="Go to Details"
          onPress={() => this.props.navigation.navigate('Details')}
        />
      </View>
    );
  }
}

Answer 10

这似乎有效：

public static navigationOptions: NavigationScreenOptionsGetter<
  NavigationScreenOptions
> = (navigation, stateProps) => ({
  title: navigation.state.params.someValue,
});

Answer 11

如果您的tsconfig.json 具有"strictNullChecks": true，则does not work 部分包含错误。在这种情况下，您确实有一个错误，因为在

行

navigation.state.params.someValue

params 是可选的。您可以做的是检查该值是否在内部传递，否则提供默认值，例如：

title: navigation.state.params && navigation.state.params.someValue || 'Default title'