如何处理PanResponder长按事件？

Question

我试图通过PanResponder处理React Native中的长按.经过一番体面的搜索,我无法找到"正确的方式",所以我在这里问.这个想法是在检测到屏幕上的长按(点击)时执行代码.我有这样的事情:

handlePanResponderGrant(e, gestureState){
    // On the press of the button set a timeout
    myVar = setTimeout(this.MyExecutableFunction(), LONG_PRESS_MIN_DURATION);
}

handlePanResponderRelease(e, gestureState) {
    // Clear the timeout if the press is released earlier than the set duration
    clearTimeout(myVar);
}

这是处理长按的正确方法还是有更好的方法？

Answer 1

我最终使用了这个功能setTimeout.这是代码,它具有检测屏幕的哪个部分被长按(左或右)的功能:

handlePanResponderGrant(e, gestureState) {
    console.log('Start of touch');

    this.long_press_timeout = setTimeout(function(){
            if (gestureState.x0 <= width/2 )
            {
                AlertIOS.alert(
                  'Left',
                  'Long click on the left side detected',
                  [
                    {text: 'Tru dat'}
                  ]
                );
            }
            else {
                AlertIOS.alert(
                  'Right',
                  'So you clicked on the right side?',
                  [
                    {text: 'Indeed'}
                  ]
                );
            }
        }, 
        LONG_PRESS_MIN_DURATION);
}
handlePanResponderMove(e, gestureState) {
    clearTimeout(this.long_press_timeout);
}
handlePanResponderRelease(e, gestureState){
    clearTimeout(this.long_press_timeout);
    console.log('Touch released');
}
handlePanResponderEnd(e, gestureState) {
    clearTimeout(this.long_press_timeout);
    console.log('Finger pulled up from the image');
}