React Native:更改ListView项目的样式 [英] React native: change style of ListView item on touch
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问题描述
我想在按下ListView项目时更新其样式,以便最终用户知道他/她选择了一个项目.
I want to update the style of a ListView item when that item is pressed so that the enduser is aware that he/she selected an item.
列表视图:
<ListView
dataSource={this.state.dataSource}
renderRow={this.renderFriend}
/>
行渲染器:
renderFriend(friend) {
return (
<TouchableHighlight onPress={ ??? }>
<View style={styles.friendItem}>
<View style={styles.profilePictureContainerNoBorder}>
<Image
source={{uri: 'https://graph.facebook.com/' + friend.id + '/picture?width=500&height=500'}}
style={styles.profilePicture}
/>
</View>
<Text style={styles.profileName}>{friend.name}</Text>
</View>
</TouchableHighlight>
);
}
当用户激活TouchableHighlight时,如何更改第二个View的样式?
How could I change the style of the second View when the user activates the TouchableHighlight?
我还想将所选对象添加到所选对象的数组中.
I would also like to add the selected object to an array of selected objects.
推荐答案
按 TouchableHighlight
时,应使用组件状态并在其中存储选定的好友ID.
You should use the component state and store the selected friends ids in it when pressing the TouchableHighlight
.
类似的东西:
constructor(props) {
super(props);
this.state = {
selectedFriendIds: [],
}
}
selectFriend(friend) {
this.setState({
selectedFriendIds: this.state.selectedFriendIds.concat([friend.id]),
});
}
renderFriend(friend) {
const isFriendSelected = this.state.selectedFriendIds.indexOf(friend.id) !== -1;
const viewStyle = isFriendSelected ?
styles.profilePictureContainerSelected : styles.profilePictureContainerNoBorder;
return (
<TouchableHighlight onPress={ () => this.selectFriend(friend) }>
<View style={styles.friendItem}>
<View style={viewStyle}>
<Image
source={{uri: 'https://graph.facebook.com/' + friend.id + '/picture?width=500&height=500'}}
style={styles.profilePicture}
/>
</View>
<Text style={styles.profileName}>{friend.name}</Text>
</View>
</TouchableHighlight>
);
}
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