[react native]onPanResponderRelease on FlatList或ListView不';android中的t触发器
我有一个FlatList组件,我想在上面实现PanResponder API,下面是我代码的一部分:[react native]onPanResponderRelease on FlatList或ListView不';android中的t触发器,android,react-native,Android,React Native,我有一个FlatList组件,我想在上面实现PanResponder API,下面是我代码的一部分: componentWillMount() { this._panResponder = PanResponder.create({ onStartShouldSetPanResponder: ()=>true, onMoveShouldSetPanResponder: ()=>true, onPanResponderGrant: (e, g
componentWillMount() {
this._panResponder = PanResponder.create({
onStartShouldSetPanResponder: ()=>true,
onMoveShouldSetPanResponder: ()=>true,
onPanResponderGrant: (e, gestureState)=>this.panResponderStart(e, gestureState),
onPanResponderMove: (e, gestureState)=>this.panResponderMove(e, gestureState),
onPanResponderRelease: (e, gestureState)=>this.panResponderEnd(e, gestureState),
onPanResponderTerminate: (e, gestureState)=>this.panResponderEnd(e, gestureState),
});
}
panResponderEnd(e, gestureState) {
//this._previousTop += gestureState.dy;
this.setState({scrollTop:0});
console.log('release');
}
...
<FlatList
{...this._panResponder.panHandlers}
...
componentWillMount(){
这是.\u panResponder=panResponder.create({
onStartShouldSetPanResponder:()=>true,
onMoveShouldSetPanResponder:()=>true,
onPanResponderGrant:(e,gestureState)=>this.panResponderStart(e,gestureState),
onPanResponderMove:(e,gestureState)=>this.panResponderMove(e,gestureState),
onPanResponderRelease:(e,gestureState)=>this.PanResponderRend(e,gestureState),
onPanResponderTerminate:(e,gestureState)=>this.panResponderEnd(e,gestureState),
});
}
PanRespondernd(e,手势状态){
//这._previousTop+=gestureState.dy;
this.setState({scrollTop:0});
console.log('release');
}
...
无需通过panHandlers
查看。我直接在部分列表中使用它。
我通过使用onpanresponderrend
而不是onPanResponderRelease
使其正常工作
此外,如果您仍要使用onPanResponderRelease
,则应通过以下方式允许终止请求:
onPanResponderTerminationRequest: () => true
如果它不起作用,请告诉我。我更改了视图上的PanResponder,例如,您可以将PanHandler
作为道具传递给平面列表
,并使其自身使用视图进行渲染。当你说它在发布前触发时,你是什么意思?是在运动中吗?在释放时,但在您的口味中太早?谢谢您的回答。当我设置乞丐查看时,它在移动期间释放