从另一个视图中关闭UIPopoverController

Question

我在另一个视图中有一个UIPopoverController命名为popover。我想知道的是，如果在当前弹出窗口视图中按下了UIButton，我该如何解除弹出窗口？提前致谢。

Answer 1

您可以使用NSNotification来告诉其他视图忽略它的弹出视图。

用法示例：

// Add an observer that will respond to our notification. 
[[NSNotificationCenter defaultCenter] addObserver:self // <- This is the object that will has the selector that we want to run (the same one we use in the next line). 
             selector:@selector(doSomething:) // <- This is the selector we want to run. 
              name:@"doSomethingNow" // <- This is notification name we will send to activate our observer's selector. 
              object:nil]; // Don't worry about this for now. 


// Post the notification. This has the same name as our observer above, so our 'doSomething' selector should be run. 
[[NSNotificationCenter defaultCenter] postNotificationName:@"doSomethingNow" object:nil]; 


// the function specified in the same class where we defined the addObserver 
- (void)doSomething:(NSNotification *)pNotification { 
    NSLog(@"Received Notification..."); 
} 

Answer 2

我一直觉得很奇怪，一个UIViewController知道它应该有多大的酥料饼通过它的“contentSizeForViewInPopover”属性，但指针不保留到UIPopoverController本身。我总是最后加入：

@property (nonatomic,assign) UIPopoverController* popover; 

我的UIViewController类，并设置，当创建popover。然后从该UIViewController中的任何东西，我可以这样做来解雇popover：

[popover dismissPopoverAnimated:YES];