UISearchDisplayController自动创建一个UIPopovercontroller来显示内容搜索结果?!怎么解雇呢? [英] UISearchDisplayController automatically creates a UIPopovercontroller to display content search result ?! How to dismiss it?
问题描述
我正在使用带有UISearchBar的UISearchDisplayController。我把这个UISearchBar放在我的应用程序中使用IB得到:
I'm using a UISearchDisplayController with a UISearchBar. I put this UISearchBar in my app using IB and I get :
alt text http://img6.imageshack.us/img6/1985/screenshot20100701at156.png
好的:当你开始录音时,结果是popovercontroller神奇地出现(我没有自己写任何东西让它出现!)
Fine : when you start taping, the result popovercontroller appears magically (I didn't write anything on my own to make it appear !)
然后,当在结果中单击一行时,我想要关闭PopoverController BUT 在这个阶段,我从未在我身边实例化过UIPopoverController:看起来如果在 UISearchDisplayController
中有一个封装的行为,它会自动包装它的 searchContentsController
在 UIPopoverController
中。这真的很棒,因为一切都运行得很好,没有做任何事情,除了我无法获得对这个UIPopoverController的引用来解雇它:(
Then, when a row is clicked among the result, I want to dismiss the PopoverController BUT at this stage, I never instantiated the UIPopoverController on my side : it looks like if there's an encapsulated behavior in the UISearchDisplayController
that automatically wraps its searchContentsController
inside a UIPopoverController
. That's really great because everything works perfectly without doing anything except that I cannot get the reference to this UIPopoverController to dismiss it :(
有谁知道如何获得引用这个神奇地创建的UIPopoverController?
(这是iPad真的是一个神奇设备的证据;)
Does anyone know how to get the reference to this "magically" created UIPopoverController ? (this is the proof the iPad is really a "magical" device ;)
我以为从它的contentController(通过它的父属性)会有一个对UIPopoverController的引用,但我找不到任何方法来获取它的指针:/
I thought there would be a reference to the UIPopoverController from its contentController (through its parent property for instance), but I cannot find any way to get a pointer to it :/
推荐答案
[searchDisplayController setActive:NO animated:YES];
不工作吗?
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