我将一个简单的React-native应用程序放在一起,从远程服务获取数据,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选择。我相信这样一个微不足道的行动应该不难。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Flatlist中突出显示项目的下一步不会发生。 'UnderlayColor'也没有任何帮助。
非常感谢任何帮助/建议。
答案 0 :(得分:6)
代替this.state.selectedItem
并使用/检查rowItem.id.value
进行设置,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 。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 prop将确保当this.state.selected在选择时发生更改时重新呈现。
你的onPressAction显然会从下面的示例中改变一点,具体取决于你是想在任何给定时间限制选择的数量,还是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;最终看起来如下:
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.get(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>
)
}
}
答案 1 :(得分:4)
您应该将extraData道具传递给FlatList,以便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/releases/next/docs/flatlist.html
确保您的renderItem函数所依赖的所有内容在更新后作为prop(例如extraData)传递,而不是===,否则您的UI可能无法更新更新
答案 2 :(得分:0)
您可以执行以下操作:
对于renderItem,请使用诸如TouchableOpacity之类的东西,并将onPress事件传递给renderItem的索引,数组的每个项目都应具有如下索引:
const CategoryList = [ {name:“一个名称”,索引:0}, .... ]
创建一个变量以控制组件的状态:
this.state = {selected_category:''}
更改上面创建的状态的功能:
_handleCategorySelect =(索引)=> { this.setState({selected_category:index}); }
FlatList:
<FlatList
data={CategoryList}
ref={(e) => this.categoryList = e}
renderItem={(category) =>
<TouchableOpacity
onPress={() => this._handleCategorySelect(category.item.index)}
style={this.state.selected_category === category.item.index ?
styles.selected : null}
>
<Text>{category.item.name}</Text>
</TouchableOpacity>
}
/>
答案 3 :(得分:0)
第一
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,我从麦孔·吉尔顿(Maicon Gilton)给出的答案中得到了解决问题的方法