我把一个简单的 React-native 应用程序放在一起,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,应突出显示该项目并保留所选内容。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项目时,将使用所选项目的信息调用"onPress"方法。但是平面列表中突出显示项的下一步不会发生。"UnderlayColor"也无济于事。
任何帮助/建议将不胜感激。
你可以执行以下操作:
-
对于
renderItem
,使用类似TouchableOpacity
的东西,其中包含传递renderedItem
索引或 id 的onPress
事件; -
将所选项目添加到状态的函数:
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})
}
-
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>
}
/>
- 为所选项目制作样式,仅此而已!
代替this.state.selectedItem
和设置/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/react-native/docs/flatlist.html。也看看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 在选择时发生更改时发生重新渲染。
您的onPressAction显然会从下面的示例中有所变化,具体取决于您是否要限制任何给定时间的选择数量或不允许用户切换选择等。
此外,虽然不是必需的,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>
)
}
}
你应该将一个额外的数据属性传递给你的 FlatList,以便它会根据你的选择重新渲染你的项目
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 renderItem 函数所依赖的所有内容在更新后都作为 === 的道具(例如 extraData)传递,否则您的 UI 可能不会在更改时更新
First
constructor() {
super();
this.state = {
selectedIds:[]
};
}
第二:
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我的问题的解决方案。