Question

在我的iPad应用程序中，我在滚动视图中使用带有箭头的弹出按钮。它工作正常，但是当我滚动视图，然后点击按钮，带箭头的弹出窗口不跟随按钮，它在原始位置打开。

我使用此代码：

(void)showHomePopupAction:(id)sender {     
   self.popHome = [[[PopHome alloc] initWithNibName:@"PopHome" bundle:[NSBundle mainBundle]] autorelease];
   popHome.contentSizeForViewInPopover = CGSizeMake(popHome.view.frame.size.width, popHome.view.frame.size.height);
   self.popoverController = [[[UIPopoverController alloc] initWithContentViewController:popHome] autorelease];
   [self.popoverController presentPopoverFromRect:popoverButtonForHome.frame inView:self.view permittedArrowDirections:UIPopoverArrowDirectionUp animated:YES];     
}

您有什么想法或建议来解决这个问题吗？谢谢！

Answer 1

嘿，试试这样的事情：

- (void)showHomePopupAction:(id)sender {     
       self.popHome = [[[PopHome alloc] initWithNibName:@"PopHome" bundle:[NSBundle mainBundle]] autorelease];
       popHome.contentSizeForViewInPopover = CGSizeMake(popHome.view.frame.size.width, popHome.view.frame.size.height);
       self.popoverController = [[[UIPopoverController alloc] initWithContentViewController:popHome] autorelease];
       CGRect frame = popoverButtonForHome.frame;
       frame.origin.y -= self.scrollView.bounds.origin.y; // you can postion the popover with + and - values
       [self.popoverController presentPopoverFromRect:frame inView:self.view permittedArrowDirections:UIPopoverArrowDirectionUp animated:YES];
}

我只需要完全一样......代码适用于uiscrollview中的uibutton。

滚动视图中的弹出窗口

1 个答案: