当按下项目时,React Native Flat列表将导航到新屏幕

问题描述 投票:1回答:2

我是React Native的新手,当我尝试构建一个平面列表时,我遇到了一个问题,当按下列表项时,该列表导航到具有项目详细信息的另一个屏幕。我正在使用redux和react-native-router-flux在屏幕之间导航。平面列表正在渲染,但是当我按下项目时,我无法导航到另一个屏幕。

这是我的代码:

class MyListItem extends React.PureComponent {

  _onPress = () => {

    this.props.onPressItem(this.props.item);

  };

  render() {

    return (
      <TouchableOpacity

        {...this.props}
        onPress={this._onPress}>
        <View style={styles.GridViewContainer} >
          <Text style={styles.GridViewTextLayout} >{this.props.item.name}</Text>
        </View>
      </TouchableOpacity>
    );
  }
}
class CustomersList extends Component {
  componentWillMount() {
    this.props.getCustomers();

  }

  componentDidUpdate(prevProps) {
    if (prevProps.customers !== this.props.customers) {
      this.props.getCustomers();
    }
  }

  _keyExtractor = (item, index) => item._id;

  _onPressItem = (item) => {
    Actions.customerShow({ customer: item });

  };

  _renderItem = ({ item }) => (
    <MyListItem
      id={item._id}
      item={item}
      onPressItem={() => this._onPressItem(item)}
      title={item.name}
    />
  );

  render = () => {
    return (
      <View style={styles.container}>
        <FlatList
          data={this.props.customers}
          keyExtractor={this._keyExtractor}
          renderItem={this._renderItem}
          extraData={this.state}
          numColumns={3}
        />
      </View>
    );
  }
}

const mapStateToProps = (state) => {
  const customers = state.customers;
  console.log(customers);
  debugger
  return { customers };

};


export default connect(mapStateToProps, {
  getCustomers
})(CustomersList);

这是axios api:

export const getCustomers = () => {
    debugger;
    return (dispatch) => {
        dispatch(setCustomerLoading)
        axios
            .get('https://calm-sands-26165.herokuapp.com/api/customers')
            .then(res =>
                dispatch({
                    type: GET_CUSTOMERS,
                    payload: res.data,
                })
            )
            .catch(err =>
                dispatch({
                    type: GET_ERRORS,
                    payload: null
                })
            );
    };
}

路由栈:

const RouterComponent = () => {
  return (
    <Router >
      <Scene key="root" hideNavBar>
        <Scene key="auth">
          <Scene key="login" component={LoginForm} title="Please Login" initial />
        </Scene>
        <Scene key="main">
          <Scene
            onRight={() => Actions.customerCreate()}
            rightTitle="Add"
            key="customersList"
            component={CustomersList}
            title="Customers"
            initial
          />
          <Scene key="customerCreate" component={CustomerCreate} title="Create Customer" />
          <Scene key="customerShow" component={CustomerShow} title="Show Customer" />
        </Scene>
      </Scene>



    </Router>
  );
};

提前致谢,

react-native redux react-native-router-flux
2个回答
1
投票

FlatList提供的数据源不是数组时,会导致错误。在您的情况下,this.props.customers未定义,直到返回函数this.props.getCustomers()

我建议使用状态来渲染CustomersList组件中的平面列表。并且当从axios异步调用this.setState({customers : nextProps.customers})返回结果时更新状态,该调用将使用customers数组重新呈现FlatList

class CustomersList extends Component {

  constructor(props){
    super(props);
      this.state = {
      customers : []
    };
  }
  ...

  render = () => {
    return (
      <View style={{flex:1}}>
        {this.state.customers.length > 0 ?
        <FlatList
          data={this.state.customers}
          keyExtractor={this._keyExtractor}
          renderItem={this._renderItem}
          extraData={this.state}
          numColumns={1}
      /> :
        <View style={styles.container}>
          <ActivityIndicator size={'large'}/>
        </View>
        }
      </View>
    );
  }
}

至于你的导航,我自己测试了你的代码并且它工作了:)(正如你所看到的,当列表仍在获取时我使用<ActivityIndicator />进行渲染。

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