当用户在UIView外部触摸时如何处理事件？

Question

I have a custom popup menu in my iOS application. It is UIView with buttons on it. How do I handle event when user touch up outside this view? I want to hide menu at this moment.

Answer 1

You should create a custom UIButton that takes up the whole screen. Then add your subview on top of that button. Then when the user taps outside of the subview they will be tapping the button.

For example, make the button like this:

UIButton * button = [UIButton buttonWithType:UIButtonTypeCustom];
[button addTarget:self action:@selector(yourhidemethod:) forControlEvents:UIControlEventTouchUpInside];
[button setTitle:@"" forState:UIControlStateNormal];
button.frame = self.view.frame;
[self.view addSubview:button];

(where yourhidemethod: is the name of the method that removes your subview.) Then add your subview on top of it.

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Update: It looks like you're wanting to know how to detect where touches are in a view. Here's what you do:

-(void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event 
{   
  UITouch *touch = [touches anyObject];
  CGPoint touchPoint = [touch locationInView:self]; //location is relative to the current view
  // do something with the touched point 
}

Answer 2

将它放在全屏透明视图中并处理它的触摸。

Answer 3

一个想法是有一个不可见的视图（uicontrol），它与屏幕一样大，然后保存这个自定义弹出窗口。

Answer 4

It sounds like you'd be better served by deriving your menu from UIControl rather than UIView. That would simplify the touch handling, and all you'd have to do in this case would be to set a target and action for UIControlEventTouchUpOutside. The target could be the menu itself, and the action would hide or dismiss the menu.