如何从 FlatList 导航到新屏幕?
2019-07-22
74
我希望在单击 FlatList 中的项目时导航到名为 GridVid 的屏幕。我不知道该如何做,因为
onPress={() => this.props.navigation.navigate('GridVid')}
只有在 App.js 中调用才有效,因为 StackNavigator 就是在那里定义的,而不是 ListItem 类(位于名为 ListItem.js 的单独文件中)
//App.js
class SettingsClass extends Component {
constructor(props) {
super(props)
this.state = {
columns: 3, //Columns for Grid
};
}
render() {
const {columns} = this.state
return (
<View style={styles.grid}>
<FlatList
numColumns={columns}
data={[
{uri:'https://randomuser.me/api/portraits/thumb/women/12.jpg'},
{uri:'https://randomuser.me/api/portraits/thumb/women/13.jpg'},
{uri:'https://randomuser.me/api/portraits/thumb/women/14.jpg'},
]}
renderItem={({item}) => {
return (<ListItem itemWidth={(ITEM_WIDTH-(10*columns))/columns}
image={item}
/>
)
}}
keyExtractor={
(index) => { return index }
}
/>
</View>
);
}
}
//Settings Class swipes to GridVid
const SettingsStack = createStackNavigator({
SettingsScreen: {
screen: SettingsClass
},
GridVid: {
screen: GridVidClass
},
});
//ListItem.js
export default class ListItem extends Component {
state = {
animatepress: new Animated.Value(1)
}
animateIn() {
Animated.timing(this.state.animatepress, {
toValue: 0.90,
duration: 200
}).start()
}
animateOut() {
Animated.timing(this.state.animatepress, {
toValue: 1,
duration: 200
}).start()
}
render() {
const {itemWidth} = this.props
return (
<TouchableWithoutFeedback
onPressIn={() => this.animateIn()}
onPressOut={() => this.animateOut()}
onPress={() => this.props.navigation.navigate('GridVid')} //WONT WORK HERE in this file!!!!
>
<Animated.View style={{
margin:5,
transform: [{scale: this.state.animatepress}] }}>
<Image style={{width:itemWidth, height: 100}} source={this.props.image}></Image>
</Animated.View>
</TouchableWithoutFeedback>
);
}
}
//GridVid.js
export default class GridVidClass extends Component {
render() {
return (
<View style={styles.container}>
<Text>On GridVid </Text>
</View>
);
}
}
有没有办法在 FlatList 中(或 App.js 中的任何位置)调用
onPress={() => this.props.navigation.navigate('GridVid')
,而不是在 ListItem 中(目前无法在那里工作)?然而在 ListItem 中,至少我单击了我想要的图像并且对我所单击的内容有一些参考。
2个回答
您需要做的是将
onPress
属性传递给您的
ListItem
,以使导航发生。
//App.js
class SettingsClass extends Component {
constructor(props) {
super(props)
this.state = {
columns: 3, //Columns for Grid
};
}
render() {
const {columns} = this.state
return (
<View style={styles.grid}>
<FlatList
numColumns={columns}
data={[
{uri:'https://randomuser.me/api/portraits/thumb/women/12.jpg'},
{uri:'https://randomuser.me/api/portraits/thumb/women/13.jpg'},
{uri:'https://randomuser.me/api/portraits/thumb/women/14.jpg'},
]}
renderItem={({item}) => {
return (<ListItem itemWidth={(ITEM_WIDTH-(10*columns))/columns}
image={item}
onPress={() => this.props.navigation.navigate('GridVid') // passing the onPress prop
/>
)
}}
keyExtractor={
(index) => { return index }
}
/>
</View>
);
}
}
//Settings Class swipes to GridVid
const SettingsStack = createStackNavigator({
SettingsScreen: {
screen: SettingsClass
},
GridVid: {
screen: GridVidClass
},
});
//ListItem.js
export default class ListItem extends Component {
state = {
animatepress: new Animated.Value(1)
}
animateIn() {
Animated.timing(this.state.animatepress, {
toValue: 0.90,
duration: 200
}).start()
}
animateOut() {
Animated.timing(this.state.animatepress, {
toValue: 1,
duration: 200
}).start()
}
render() {
const {itemWidth} = this.props
return (
<TouchableWithoutFeedback
onPressIn={() => this.animateIn()}
onPressOut={() => this.animateOut()}
onPress={this.props.onPress} // using onPress prop to navigate
>
<Animated.View style={{
margin:5,
transform: [{scale: this.state.animatepress}] }}>
<Image style={{width:itemWidth, height: 100}} source={this.props.image}></Image>
</Animated.View>
</TouchableWithoutFeedback>
);
}
}
Vencovsky
2019-07-22
ListItem 不在 StackNavigator 中,所以它不知道什么是导航
你可以像 Vencovsky 的答案那样去做,或者从 ListItem 的父组件传递导航属性
<ListItem
itemWidth={(ITEM_WIDTH-(10*columns))/columns}
image={item}
navigation={this.props.navigation}
/>
cuongtd
2019-07-22