Question

我在iPad上为image picker创建了一个弹出框。这个功能有效。但如果按两次UIBarButton，应用程序崩溃。

@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];
}

你有什么建议吗？

Answer 1

您可能需要检查弹出窗口是否可见。如果您已创建其对象，则将其关闭并再次创建popover

if ([self.popoverController isPopoverVisible]) {
    [self.popoverController dismissPopoverAnimated:YES];
    [popoverController setDelegate:nil];
    [popoverController release]; // Use release only if , it is without ARC
}
else
{
    // Create popover and assign its properties.
}

这肯定会解决您在所有iOS版本中的问题。 :)有一个快乐的编码。!!

iPad iOS 6.1 UIPopover ImagePicker在按下两次条形按钮时崩溃

1 个答案: