React-native - 将数据从一个屏幕传递到另一个屏幕

Question

我正在尝试学习如何使用react-native,所以我建立了一个小应用程序,从服务器获取用户列表,将其显示在一个listview按钮中,打开一个显示用户数据的屏幕.

我设置了一个导航器,从一个屏幕到另一个屏幕.按下行时,我可以打开一个新屏幕,但我无法想象如何将数据传递到这个新屏幕.

我在里面设置了一个导航仪 index.android.js

import React from 'react';
import {
  AppRegistry,
  Navigator,
} from 'react-native';

import ContactList from './src/containers/ContactList.js';

function MyIndex() {
  return (
    <Navigator
      initialRoute={{ name: 'index', component: ContactList }}
      renderScene={(route, navigator) => {
        if (route.component) {
          return React.createElement(route.component, { navigator });
        }

        return undefined;
      }}
    />
  );
}

AppRegistry.registerComponent('reactest', () => MyIndex);

首先,我显示一个带有按钮和空列表视图的屏幕(ContactList.js),按下按钮后,我获取了一些用于更新的JSON数据listview:

import React, { Component, PropTypes } from 'react';
import {
  Text,
  View,
  TouchableOpacity,
  TouchableHighlight,
  ListView,
  Image,
} from 'react-native';

import styles from '../../styles';
import ContactDetails from './ContactDetails';

const url = 'http://api.randomuser.me/?results=15&seed=azer';

export default class ContactList extends Component {
  static propTypes = {
    navigator: PropTypes.object.isRequired,
  }
  constructor(props) {
    super(props);

    const datasource = new ListView.DataSource({ rowHasChanged: (r1, r2) => r1 !== r2 });
    this.state = {
      jsonData: datasource.cloneWithRows([]),
      ds: datasource,
    };
  }
  _handlePress() {
    return fetch(url)
      // convert to json
      .then((response) => response.json())
      // do some string manipulation on json
      .then(({ results }) => {
        const newResults = results.map((user) => {
          const newUser = {
            ...user,
            name: {
              title: `${user.name.title.charAt(0).toUpperCase()}${user.name.title.slice(1)}`,
              first: `${user.name.first.charAt(0).toUpperCase()}${user.name.first.slice(1)}`,
              last: `${user.name.last.charAt(0).toUpperCase()}${user.name.last.slice(1)}`,
            },
          };

          return newUser;
        });

        return newResults;
      })
      // set state
      .then((results) => {
        this.setState({
          jsonData: this.state.ds.cloneWithRows(results),
        });
      });
  }
  renderRow(rowData: string) {
    return (
      <TouchableHighlight
        onPress={() => {
          this.props.navigator.push({
            component: ContactDetails,
          });
        }}
      >
        <View style={styles.listview_row}>
          <Image
            source={{ uri: rowData.picture.thumbnail }}
            style={{ height: 48, width: 48 }}
          />
          <Text>
            {rowData.name.title} {rowData.name.first} {rowData.name.last}
          </Text>
        </View>
      </TouchableHighlight>
    );
  }
  render() {
    const view = (
      <View style={styles.container}>
        <TouchableOpacity
          onPress={() => this._handlePress()}
          style={styles.button}
        >
          <Text>Fetch results?</Text>
        </TouchableOpacity>
        <ListView
          enableEmptySections
          dataSource={this.state.jsonData}
          renderRow={(rowData) => this.renderRow(rowData)}
          onPress={() => this._handleRowClick()}
        />
      </View>
    );

    return view;
  }
}

按下一行时,它会打开一个新屏幕ContactDetails.js,该屏幕应显示用户的数据:

import React, {
} from 'react';

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

import styles from '../../styles';

export default function ContactDetails() {
  return (
    <View style={styles.container}>
      <Text>{this.props.title}</Text>
      <Text>{this.props.first}</Text>
      <Text>{this.props.last}</Text>
    </View>
  );
}

此时我收到了这个错误:

undefined不是对象(评估'this.props.title')

我尝试了很多东西,比如:

this.props.navigator.push({
    component: ContactDetails,
    title: rowData.name.title,
    first: rowData.name.first,
    last: rowData.name.last,
});

要么

this.props.navigator.push({
    component: ContactDetails,
    props: {
        title: rowData.name.title,
        first: rowData.name.first,
        last: rowData.name.last,
    }
});

要么

this.props.navigator.push({
    component: ContactDetails,
    passProps: {
        title: rowData.name.title,
        first: rowData.name.first,
        last: rowData.name.last,
    }
});

但无济于事.

我还读到我应该使用redux.难道我做错了什么 ？

编辑:问题在这里:

<TouchableHighlight
    onPress={() => {
      this.props.navigator.push({
        component: ContactDetails,
      });
    }}
>

我想我应该在那里传递一些参数,但无论我在上面尝试过什么都失败了.

Answer 1

所以问题似乎在于:

<Navigator
  initialRoute={{ name: 'index', component: ContactList }}
  renderScene={(route, navigator) => {
    if (route.component) {
      return React.createElement(route.component, { navigator });
    }

    return undefined;
  }}
/>

在renderScene函数中,您确实收到了路由(并使用route.component),但是您没有传递route.props或route.passProps或者您想要调用它的内容!从我在Navigator的源代码中看到的那一刻起,你应该在创建它时拥有完整的路径对象.所以你应该能够传递你的道具.

例如:

<Navigator
  initialRoute={{ name: 'index', component: ContactList }}
  renderScene={(route, navigator) => {
    if (route.component) {
      return React.createElement(route.component, { navigator, ...route.props });
    }

    return undefined;
  }}
/>

// push
<TouchableHighlight
  onPress={() => {
    this.props.navigator.push({
      component: ContactDetails,
      props: { /* ... */ }
    });
  }}
>

您也可以设置redux,但这不是必需的.虽然如果您的应用程序变大,您应该考虑使用外部商店!

更新:

还有另一个问题.

您使用功能组件.功能组件没有this.他们在参数中接收道具.

所以它应该是这样的:

export default function ContactDetails(props) {
  return (
    <View style={styles.container}>
      <Text>{props.title}</Text>
      <Text>{props.first}</Text>
      <Text>{props.last}</Text>
    </View>
  );
}