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我在iPad上爲image picker
創建了一個彈出窗口。此功能起作用。但是,如果您按兩次UIBarButton
,則應用程序崩潰。iPad iOS 6.1 UIPopover ImagePicker在按兩次條形按鈕時崩潰
@property (retain) UIPopoverController *popoverController1;
-(IBAction)photos:(id)sender {
test = false;
UIImagePickerController *imagePicker = [[UIImagePickerController alloc] init];
imagePicker.wantsFullScreenLayout = NO;
imagePicker.delegate = self;
imagePicker.sourceType = UIImagePickerControllerSourceTypePhotoLibrary;
imagePicker.allowsEditing = YES;
self.popoverController1 = [[UIPopoverController alloc] initWithContentViewController:imagePicker];
_popoverController1.delegate = self;
[_popoverController1 setPopoverContentSize:CGSizeMake(1024, 500)];
[self.popoverController1 presentPopoverFromBarButtonItem:sender permittedArrowDirections:UIPopoverArrowDirectionAny animated:YES];
}
您有什麼建議嗎?
什麼是崩潰的錯誤? – rmaddy 2013-03-04 23:38:55
正確的做法是檢查點擊按鈕時是否顯示彈出窗口。如果是這樣,請關閉當前的彈出窗口。如果不是,則顯示彈出窗口。 – rmaddy 2013-03-04 23:39:51
這是錯誤:終止應用程序,由於未捕獲的異常'NSGenericException',原因:' - [UIPopoverController dealloc]達到popover仍然可見。' 這是一個很好的建議。謝謝! :D我該如何檢查? – 2013-03-04 23:47:24