如何将这些数据返回到其他页面反应导航页面

Question

import React from 'react';
import { View, Text, Button, Alert } from 'react-native';
import Expo from 'expo';
export default class LoginScreen extends React.Component{
async signInWithGoogleAsync() {
    try {
      const result = await Expo.Google.logInAsync({
        androidClientId: '768673945897-ftrhehlg3io4v2fohdbpacio1vghos8v.apps.googleusercontent.com',
        //iosClientId: YOUR_CLIENT_ID_HERE,
        scopes: ['profile', 'email'],
      });
      if (result.type === 'success') {
            console.log(result);
            //this.props.navigation.navigate('Profile');
            return result.accessToken;
            return result.user.name;
            return result.user.photoUrl;
      } else {
        return {cancelled: true};
      }
    } catch(e) {
      return {error: true};
    }
  }
render() {
    return(
        <View>
            <Button 
                onPress={this.signInWithGoogleAsync.bind(this)}
                title='Google Sign in'     
            />
        </View>
    );
}
}

Answer 1

如果你想访问其他组件或页面中的数据，你应该使用redux存储。您可以将在当前页面中获取或生成的数据存储到 redux 存储中，任何需要它的组件或页面都可以访问该存储。

Answer 2

您可以尝试将参数传递给这样的路由

class HomeScreen extends React.Component {
  render() {
    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 */
            this.props.navigation.navigate('Details', {
              itemId: 86,
              otherParam: 'anything you want here',
            });
          }}
        />
      </View>
    );
  }
}

并像这样检索数据...

class DetailsScreen extends React.Component {
  render() {
    /* 2. Get the param, provide a fallback value if not available */
    const { navigation } = this.props;
    const itemId = navigation.getParam('itemId', 'NO-ID');
    const otherParam = navigation.getParam('otherParam', 'some default value');
    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>
        <Button
          title="Go to Details... again"
          onPress={() =>
            this.props.navigation.push('Details', {
              itemId: Math.floor(Math.random() * 100),
            })}
        />
        <Button
          title="Go to Home"
          onPress={() => this.props.navigation.navigate('Home')}
        />
        <Button
          title="Go back"
          onPress={() => this.props.navigation.goBack()}
        />
      </View>
    );
  }
}

Answer 3

1. 它已经解决了，感谢您的帮助:)）顺便说一句，我只是将此数组数据传递到导航器:)

结果 = [对象]

this.props.navigation.navigate（'Profile'， { result： result.user }）;

2. 然后在其他页面中，让我们说个人资料，我为新状态制作了

this.state = { result： this.props.navigation.state.params.result， };

3. 然后我在每个部分中都像这样称呼该状态

{ this.state.result.email }