ios – 如何检测在UIImageView外部点击的人

当有人在应用程序的任何部分点击UIImageView之外时,我希望UIImageView能够消失.

@interface SomeMasterViewController : UITableViewController <clip>

<clip>

@property (strong,nonatomiC) UIImageView *someImageView;

stackoverflow和apple的文档中有一些提示听起来像我需要的.

> Apple’s : Gesture Recognizers
> Apple’s : UIView hitTest:withEvent
> Apple’s : UITouch Class Reference
> Stackoverflow: Listening to UITouch event along with UIGestureRecognizer
> (not likely needed but..) – CGRectContainsPoint as mentioned in the following post titled: Comparing a UITouch location to UIImageView rectangle

但是,我想在这里查看我的方法.我的理解是代码需要

>注册UITapGestureRecognizer以获取应用程序中可能发生的所有触摸事件
> UITapGestureRecognizer应该有cancelsTouchesInView和
delaysTouchesBegan和delaysTouchesEnded设置为NO.
>将这些触摸事件与someImageView进行比较(如何使用UIView hitTest：withEvent？)

更新：我正在使用主UIWindow注册UITapGestureRecognizer.

最终未解决的部分

我有一个UTapGestureRecognizer将调用的handleTap：(UITapGestureRecognizer *).如何获取给定的UITapGestureRecognizer并查看水龙头是否落在UIImageView之外？识别器的LOCATIOnInView看起来很有希望,但我没有得到我期望的结果.当我点击它时我希望看到某个UIImageView,当我点击另一个点时看不到UIImageView.我觉得LOCATIOnInView方法使用错误.

这是我对LOCATIOnInView方法的调用：

- (void)handleTap:(UITapGestureRecognizer *)gestureRecognizer
{
    if (gestureRecognizer.state != UIGestureRecognizerStateEnded) {
        NSLog(@"handleTap NOT given UIGestureRecognizerStateEnded so nothing more to do");
        return;        
    }

    UIWindow *mainWindow = [[[UIApplication sharedApplication] delegate] window];
    CGPoint point = [gestureRecognizer LOCATIOnInView:mainWindow];
    NSLog(@"point x,y computed as the LOCATIOn in a given view is %f %f",point.x,point.y);

    UIView *touchedView = [mainWindow hitTest:point withEvent:nil];
    NSLog(@"touchedView = %@",touchedView); 
}

我得到以下输出：

<clip>point x,y computed as the LOCATIOn in a given view is 0.000000 0.000000

<clip>touchedView = <UIWindow: 0x8c4e530; frame = (0 0; 768 1024); opaque = NO; autoresize = RM+BM; layer = <UIWindowLayer: 0x8c4c940>>

谢谢！