React native React Native中FlatList内的SlidengUpPanel按钮
我正在努力解决SlidingUpPanel和FlatList的问题。 我想做的是: 包含React native React Native中FlatList内的SlidengUpPanel按钮,react-native,react-native-android,react-native-flatlist,React Native,React Native Android,React Native Flatlist,我正在努力解决SlidingUpPanel和FlatList的问题。 我想做的是: 包含 Obj 1 Obj 2 Obj 3 如果我按每个对象1或3,我想弹出幻灯片面板 我设计的: <SafeareaView> <View style={myStyle}> <SlidingUpPanel> ref={c => (_panel = c)} ... </Slidin
<SafeareaView>
<View style={myStyle}>
<SlidingUpPanel>
ref={c => (_panel = c)}
...
</SlidingUpPanel>
</View>
<FlatList>
<TouchableOpacity onPress={() => _panel.show(height);}>
...
</TouchableOpacity>
</FlatList>
</SafeAreaView>
ref={c=>(_panel=c)}
...
_面板显示(高度);}>
...
但它不起作用,我挣扎了很长时间。事实上,我不确定FlatList中的“onPress\u panel.show”是否可以指向SlidengAppAnel中的“\u panel” 我错过什么了吗?是否有关于SlidengAppanel和FlatList的建议或示例?
感谢您的帮助。注意:此处和此处可能存在问题
ref={c=>(_panel=c)}
...
(
_面板显示(高度);}>
...
)}
keyExtractor={item=>item.id.toString()}
/>
将ScrollView或FlatList包装到另一个视图中,并为该视图指定高度。像这样
<SafeareaView>
<View style={myStyle}>
<SlidingUpPanel>
ref={c => (_panel = c)}
...
<View style={{ height: 236 }}>
<FlatList
data={this.state.locations}
renderItem={({ item }) => (
<TouchableOpacity onPress={() => _panel.show(height);}>
...
</TouchableOpacity>
)}
keyExtractor={item => item.id.toString()}
/>
</View>
</SlidingUpPanel>
</View>
</SafeAreaView>