在scrollview iOS中弹出窗口的错误位置

Question

我有一个显示弹出窗口的按钮。弹出窗口必须出现在按钮下方。但它出现在屏幕上的某个地方。

这是弹出窗口的我的代码

UIView *popoverView = [[UIView alloc]initWithFrame:CGRectMake(0, 0, 320,340)]; 
[popoverView addSubview:tblViewMenu];
popoverContent.view = popoverView;
popoverContent .contentSizeForViewInPopover = CGSizeMake(620,620);
popPickerController = [[UIPopoverController alloc]initWithContentViewController:popoverContent];
[popPickerController presentPopoverFromRect:anchor.frame inView:self.view permittedArrowDirections:UIPopoverArrowDirectionUpanimated:YES];

顺便说一下，这个popover会显示在一个滚动视图中。

Answer 1

将以下代码用于当前的popoverView。

[popover presentPopoverFromRect:button.frame inView:self.scrollView permittedArrowDirections: UIPopoverArrowDirectionLeft | UIPopoverArrowDirectionUp animated:YES];

Answer 2

将按钮的直接父级作为参数值inView:

[popPickerController presentPopoverFromRect:anchor.frame
                                     inView:anchor.superview
                   permittedArrowDirections:UIPopoverArrowDirectionUp
                                   animated:YES];

始终确保在呈现popover时给予正确的父母。

Answer 3

[ur popovername presentPopoverFromRect：[（UIButton *）sender frame] inView：anchor.superview allowedArrowDirections：UIPopoverArrowDirectionUp animated：YES];