使用React-Native Navigation传递数据

wdl*_*x11 13 javascript android ios reactjs react-native

我试图在我的应用程序中的屏幕之间传递数据.目前我正在使用

"react-native": "0.46.0",
"react-navigation": "^1.0.0-beta.11"
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我有我的index.ios

 import React, { Component } from 'react';
    import {
      AppRegistry,
    } from 'react-native';
    import App from './src/App'
    import { StackNavigator } from 'react-navigation';
    import SecondScreen from './src/SecondScreen'    

    class med extends Component {
      static navigationOptions = {
        title: 'Home Screen',
      };

      render(){
        const { navigation } = this.props;

        return (
          <App navigation={ navigation }/>
        );
      }
    }

    const SimpleApp = StackNavigator({
      Home: { screen: med },
      SecondScreen: { screen: SecondScreen, title: 'ss' },    
    });

    AppRegistry.registerComponent('med', () => SimpleApp);
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应用为

    import React, { Component } from 'react';
    import {
      StyleSheet,
      Text,
      Button,
      View
    } from 'react-native';
    import { StackNavigator } from 'react-navigation';

    const App = (props)  => {
      const { navigate } = props.navigation;

      return (
        <View>
          <Text>
            Welcome to React Native Navigation Sample!
          </Text>
          <Button
              onPress={() => navigate('SecondScreen', { user: 'Lucy' })}
              title="Go to Second Screen"
            />
        </View>
      );
    }

    export default App
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然后秒屏

    import React, { Component } from 'react';
    import {
      StyleSheet,
      Text,
      View,
      Button
    } from 'react-native';

    import { StackNavigator } from 'react-navigation';


    const SecondScreen = (props)  => {
      const { state} = props.navigation;
      console.log("PROPS" + state.params);


      return (
        <View>
          <Text>
            HI
          </Text>

        </View>
      );
    }

    SecondScreen.navigationOptions = {
      title: 'Second Screen Title',
    };

    export default SecondScreen

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每当我调试console.log时,我都会被定义.
https://reactnavigation.org/docs/navigators/navigation-prop 文档说每个屏幕应该有这些值我做错了什么?

Ped*_*lin 16

所有其他答案现在似乎已经过时。在当前的 React 导航版本 ( "@react-navigation/native": "^5.0.8",) 中,您首先在一个屏幕之间传递值,如下所示:

       function HomeScreen({ navigation }) {
      return (
        <View style={{ flex: 1, alignItems: 'center', justifyContent: 'center' }}>
          <Text>Home Screen</Text>
          <Button
            title="Go to Details"
            onPress={() => {
              /* 1. Navigate to the Details route with params, passing the params as an object in the method navigate */
              navigation.navigate('Details', {
                itemId: 86,
                otherParam: 'anything you want here',
              });
            }}
          />
        </View>
      );
    }
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然后在您正在重定向的组件中,您将获得这样传递的数据:

function DetailsScreen({ route, navigation }) {
  /* 2. Get the param */
  const { itemId } = route.params;
  const { otherParam } = route.params;
  return (
    <View style={{ flex: 1, alignItems: 'center', justifyContent: 'center' }}>
      <Text>Details Screen</Text>
      <Text>itemId: {JSON.stringify(itemId)}</Text>
      <Text>otherParam: {JSON.stringify(otherParam)}</Text>
    </View>
  );
}
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所以,基本上,数据现在在里面this.props.route.params。在上面的那些例子中,我展示了如何从功能组件中获取它们,但在类组件中是类似的,我做了这样的事情:

首先我从这个 ProfileButton 传递数据,在它的handleNavigate函数中,像这样:


// these ProfileButton and ProfileButtonText, are a Button and a Text, respectively,
// they were just styled with styled-components 
<ProfileButton
 onPress={() => this.handleNavigate(item) 
  <ProfileButtonText>
      check profile
  </ProfileButtonText>
</ProfileButton>
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哪里handleNavigate是这样的:

   handleNavigate = user => {
        // the same way that the data is passed in props.route,
        // the navigation and it's method to navigate is passed in the props.
        const {navigation} = this.props;
        navigation.navigate('User', {user});
    };

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然后,函数 HandleNavigate 重定向到用户页面,这是一个类组件,我得到这样的数据:

import React, {Component} from 'react';
import {View, Text} from 'react-native';

export default class User extends Component {
    state = {
        title: this.props.route.params.user.name,
    };


    render() {
        const {title} = this.state;
        return (
            <View>
                <Text>{title}</Text>
            </View>
        );
    }
}

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在类组件中,我发现的方法是使这条线很长, title: this.props.route.params.user.name,但它有效。如果有人知道如何在当前版本的 react-native 导航中缩短它,请赐教。我希望这能解决你的问题。


Pop*_*ack 15

在你的代码中,props.navigation和this.props.navigation.state是两个不同的东西.您应该在第二个屏幕中尝试这个:

const {state} = props.navigation;
console.log("PROPS " + state.params.user);
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const {state}条线只是为了获得易于阅读的代码.


小智 9

头等舱

<Button onPress = {
  () => navigate("ScreenName", {name:'Jane'})
} />
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二等

const {params} = this.props.navigation.state
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小智 5

您可以访问您的PARAM是user,与props.navigation.state.params.user相关组件(SecondScreen)。


Luc*_*nte 5

反应导航3. *

家长班

this.props.navigation.navigate('Child', {
    something: 'Some Value',
});
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儿童班

this.props.navigation.state.params.something // outputs "Some Value"
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