突出显示 React-Native FlatList 中的选定项目

问题描述 投票:0回答:4

我编写了一个简单的 React-native 应用程序来从远程服务获取数据,并将其加载到 FlatList 中。当用户点击某个项目时,该项目应突出显示并保留选择。我相信这样一个微不足道的操作应该不难。我不确定我错过了什么。

import React, { Component } from 'react';
import {
  StyleSheet,
  Text,
  View,
  FlatList,
  ActivityIndicator,
  Image,
  TouchableOpacity,
} from 'react-native';

export default class BasicFlatList extends Component {
  constructor(props) {
    super(props);

    this.state = {
      loading: false,
      data: [],
      page: 1,
      seed: 1,
      error: null,
      refreshing: false,
      selectedItem:'null',
    };
  }

  componentDidMount() {
    this.makeRemoteRequest();
  }

  makeRemoteRequest = () => {
    const {page, seed} = this.state;
    const url = `https://randomuser.me/api/?seed=${seed}&page=${page}&results=20`;
    this.setState({loading: true});
    fetch(url)
      .then(res => res.json())
      .then(res => {
        this.setState({
          data: page === 1 ? res.results : [...this.state.data, ...res.results],
          error: res.error || null,
          loading: false,
          refreshing: false
        });
      })
      .catch(error => {
        this.setState({error, loading: false});
      });
  };

  onPressAction = (rowItem) => {
    console.log('ListItem was selected');
    console.dir(rowItem);
    this.setState({
      selectedItem: rowItem.id.value
    });
  }

  renderRow = (item) => {
    const isSelectedUser = this.state.selectedItem === item.id.value;
    console.log(`Rendered item - ${item.id.value} for ${isSelectedUser}`);
    const viewStyle = isSelectedUser ? styles.selectedButton : styles.normalButton;
    return(
        <TouchableOpacity style={viewStyle} onPress={() => this.onPressAction(item)} underlayColor='#dddddd'>
          <View style={styles.listItemContainer}>
            <View>
              <Image source={{ uri: item.picture.large}} style={styles.photo} />
            </View>
            <View style={{flexDirection: 'column'}}>
              <View style={{flexDirection: 'row', alignItems: 'flex-start',}}>
                {isSelectedUser ?
                  <Text style={styles.selectedText}>{item.name.first} {item.name.last}</Text>
                  : <Text style={styles.text}>{item.name.first} {item.name.last}</Text>
                }
              </View>
              <View style={{flexDirection: 'row', alignItems: 'flex-start',}}>
                <Text style={styles.text}>{item.email}</Text>
              </View>
            </View>
          </View>
        </TouchableOpacity>
    );
  }

  render() {
    return(
      <FlatList style={styles.container}
        data={this.state.data}
        renderItem={({ item }) => (
          this.renderRow(item)
        )}
      />
    );
  }
}

const styles = StyleSheet.create({
  container: {
    flex: 1,
    marginTop: 50,
  },
  selectedButton: {
    backgroundColor: 'lightgray',
  },
  normalButton: {
    backgroundColor: 'white',
  },
  listItemContainer: {
    flex: 1,
    padding: 12,
    flexDirection: 'row',
    alignItems: 'flex-start',
  },
  text: {
    marginLeft: 12,
    fontSize: 16,
  },
  selectedText: {
    marginLeft: 12,
    fontSize: 20,
  },
  photo: {
    height: 40,
    width: 40,
    borderRadius: 20,
  },
});

当用户点击列表中的某个项目时,将使用所选项目的信息调用“onPress”方法。但是 Flatlist 中突出显示项目的下一步不会发生。 “UnderlayColor”也没有帮助。

任何帮助/建议将不胜感激。

react-native react-native-flatlist
4个回答
19
投票

你可以这样做:

  1. 对于

    renderItem
    ,使用类似
    TouchableOpacity
    的东西,并通过
    onPress
    事件传递
    renderedItem
    的索引或 id;

  2. 将所选项目添加到状态的功能:

handleSelection = (id) => {
   var selectedId = this.state.selectedId

   if(selectedId === id)
     this.setState({selectedItem: null})
   else
     this.setState({selectedItem: id})
}

handleSelectionMultiple = (id) => {
   var selectedIds = [...this.state.selectedIds] // clone state

   if(selectedIds.includes(id))
     selectedIds = selectedIds.filter(_id => _id !== id)
   else
     selectedIds.push(id)

   this.setState({selectedIds})
}
  1. FlatList
<FlatList
  data={data}
  extraData={
    this.state.selectedId     // for single item
    this.state.selectedIds    // for multiple items
  }
  renderItem={(item) =>
     <TouchableOpacity

       // for single item
       onPress={() => this.handleSelection(item.id)}
       style={item.id === this.state.selectedId ? styles.selected : null}

       // for multiple items
       onPress={() => this.handleSelectionMultiple(item.id)}
       style={this.state.selectedIds.includes(item.id) ? styles.selected : null}
     >
        <Text>{item.name}</Text>
     </TouchableOpacity>
  }
/>
  1. 为所选项目创建样式即可!

11
投票

代替

this.state.selectedItem
并设置/检查
rowItem.id.value
,我建议使用带有键:值对的 Map 对象,如 RN FlatList 文档示例所示:https://facebook.github.io/反应本机/docs/flatlist.html。另请查看 js Map 文档:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map

@j.I-V 推荐的 extraData 属性将确保当 this.state.selected 在选择上发生更改时发生重新渲染。

您的 onPressAction 显然会与下面的示例有所不同,具体取决于您是否想限制在任何给定时间的选择数量或不允许用户切换选择等。

此外,虽然没有必要,但我喜欢为

renderItem
组件使用另一个类或纯组件;最终看起来像下面这样:

export default class BasicFlatList extends Component {
  state = {
    otherStateStuff: ...,
    selected: (new Map(): Map<string, boolean>) //iterable object with string:boolean key:value pairs
  }

  onPressAction = (key: string) => {
    this.setState((state) => {
      //create new Map object, maintaining state immutability
      const selected = new Map(state.selected);
      //remove key if selected, add key if not selected
      this.state.selected.has(key) ? selected.delete(key) : selected.set(key, !selected.get(key));
      return {selected};
    });
  }

  renderRow = (item) => {
    return (
        <RowItem
          {...otherProps}
          item={item}
          onPressItem={this.onPressAction}
          selected={!!this.state.selected.get(item.key)} />
    );
  }

  render() {
    return(
      <FlatList style={styles.container}
        data={this.state.data}
        renderItem={({ item }) => (
          this.renderRow(item)
        )}
        extraData={this.state}
      />
    );
  }
}


class RowItem extends Component {
  render(){
    //render styles and components conditionally using this.props.selected ? _ : _
    
    return (
      <TouchableOpacity onPress={this.props.onPressItem}>
        ...
      </TouchableOpacity>
    )
  }
}


7
投票

您应该将 extraData 属性传递给您的 FlatList,以便它根据您的选择重新渲染您的项目

这里:

<FlatList style={styles.container}
    data={this.state.data}
    extraData={this.state.selectedItem}
    renderItem={({ item }) => (
      this.renderRow(item)
    )}
 />

来源:https://facebook.github.io/react-native/docs/flatlist

确保您的 renderItem 函数所依赖的所有内容都作为 prop(例如 extraData)传递,并且在更新后不是 === ,否则您的 UI 可能不会因更改而更新


0
投票

第一

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

第二个

handleSelectionMultiple = async (id) => {
var selectedIds = [...this.state.selectedIds] // clone state
if(selectedIds.includes(id))
  selectedIds = selectedIds.filter(_id => _id !== id)
else 
  selectedIds.push(id)
await this.setState({selectedIds})
}

第三

<CheckBox
     checked={this.state.selectedIds.includes(item.expense_detail_id) ? true : false}
     onPress={()=>this.handleSelectionMultiple(item.expense_detail_id)}
 />

最后我从 Maicon Gilton 给出的答案中得到了问题的解决方案

© www.soinside.com 2019 - 2024. All rights reserved.