React Native <listitem onpress =“{...}”> - 为什么这个函数在执行?

时间:2016-10-27 02:29:38

标签: javascript react-native react-native-router-flux native-base

我刚刚开始围绕React Native,并使用Redux管理state,我的UI组件NativeBase libraryreact-native-router-flux处理视图之间的导航。

目前我正在构建一个基于guest个对象数组创建的基本列表。该列表存储在Redux存储中,我可以访问它并相应地显示。列表中的每个项目都与guest数组中的guests相关联。我想让列表中的每个项目都可触摸,然后将相关的guest对象作为属性传递给下一个视图以显示详细信息。

为此,我使用onPress函数,该函数是与ListItem组件关联的标准函数(请参阅NativeBase docs here)。我还为react-native-router-flux跟踪了the documentation,并在组件本身之外定义了导航操作,以便ListItem在按下时调用它而不是每次渲染时调用它。

我的问题是,我发现每次呈现onPress={goToView2(guest)}时都会调用ListItem函数,而不是在按下ListItem时调用。{/ p>

我确信它一定是简单的,我省略了。有什么建议吗?

View1.js - 显示guests的初始列表的视图:

import React, { Component } from 'react';
import { Container, Header, Title, Content, Footer, FooterTab, Button, Icon,
  Text, List, ListItem } from 'native-base';
import { connect } from 'react-redux';
import { Actions as NavigationActions } from 'react-native-router-flux';

class View1 extends Component {
  render() {

    // This is the function that runs every time a ListItem component is rendered. 
    // Why is this not only running on onPress?

    const goToView2 = (guest) => {
      NavigationActions.view2(guest);
      console.log('Navigation router run...');
    };

    return (
      <Container>
        <Header>
          <Title>Header</Title>
        </Header>


        <Content>
          <List
            dataArray={this.props.guests}
            renderRow={(guest) =>
              <ListItem button onPress={goToView2(guest)}>
                <Text>{guest.name}</Text>
                <Text note>{guest.email}</Text>
              </ListItem>
            }
          >
          </List>
        </Content>

        <Footer>
          <FooterTab>
            <Button transparent>
              <Icon name='ios-call' />
            </Button>
          </FooterTab>
        </Footer>
      </Container>
    );
  }
}

const mapStateToProps = state => {
  console.log('mapStateToProps state', state);
  return { guests: state.guests };
};

export default connect(mapStateToProps)(View1);

View2.js - 显示guest中所选View1.js的详细信息的视图:

import React, { Component } from 'react';
import { Container, Header, Title, Content, Footer, FooterTab, Button, Icon,
  Text, List, ListItem } from 'native-base';

class View2 extends Component {
    render() {
        console.log('View2 props: ', this.props);

        return (
            <Container>
                <Header>
                    <Title>Header</Title>
                </Header>

                <Content>
                <Content>
                <List>
                    <ListItem>
                        <Text>{this.props.name}</Text>
                    </ListItem>
                </List>
            </Content>
                </Content>

                <Footer>
                    <FooterTab>
                        <Button transparent>
                            <Icon name='ios-call' />
                        </Button>
                    </FooterTab>
                </Footer>
            </Container>
        );
    }
}

export default View2;

2 个答案:

答案 0 :(得分:33)

尝试<ListItem button onPress={() => {goToView2(guest)}}

答案 1 :(得分:2)

const goToView2 = (guest) => {
      NavigationActions.view2(guest);
      console.log('Navigation router run...');
    };

<ListItem button onPress={this.goToView2(guest)}>

如果你在函数中使用箭头函数那么它只是this.goToView2(来宾)...在渲染方法中写入箭头函数会导致性能问题......即使它可以忽略不计,最好避免它