长按手势打开弹出窗口，如 Whatsapp

Question

我想在长按手势时打开弹出窗口。我的应用程序有 UITableView，当用户长按 UITableviewCell 打开弹出窗口时。当用户按住他的手指足够长时，则只显示弹出窗口。当用户长按并松开手指时不会。

我正在使用以下代码：当我松开手指时使用此代码，后弹出窗口将打开，所以这是错误的。我想在不松开手指的情况下长按打开弹出窗口。

//Long press gesture
UILongPressGestureRecognizer *longPressGesture= [[UILongPressGestureRecognizer alloc]initWithTarget:self action:@selector(handleLongPress:)];
longPressGesture.minimumPressDuration = .4; //seconds
longPressGesture.delegate = self;
longPressGesture.delaysTouchesBegan = YES;
cell.titleLabel.userInteractionEnabled = YES;
[cell.titleLabel addGestureRecognizer:longPressGesture];

Answer 1

如果你想在长按开始时立即采取一些行动，那么你必须检查状态是否为 UIGestureRecognizerStateBegan 然后编写你想要在长按手势开始时执行的代码。尝试使用以下代码行。

 -(void) handleLongPress:(UILongPressGestureRecognizer *)sender
{
  if (sender.state == UIGestureRecognizerStateBegan)
  {
  //Write code for open pop up.
  }
}

Answer 2

你可以这样做：

 -(void) handleLongPress:(UILongPressGestureRecognizer *)sender
{
  if (sender.state == UIGestureRecognizerStateBegan)
  {
  //Start a timer and perform action after whatever time interval you want.
  }
  if (sender.state == UIGestureRecognizerStateEnded)
  {
  //Check the duration and if it is less than what you wanted, invalidate the timer.
  }
}

长按手势打开弹出窗口，如 Whatsapp

Open popup on long press gesture like Whats app

objective-c

popup

ios

uilongpressgesturerecogni