滾動視圖中的彈出式對話框

Question

在我的iPad應用程序中，我使用滾動視圖中的按鈕上帶有箭頭的彈出式對話框。它工作正常，但當我滾動視圖時，然後點擊按鈕，帶箭頭的彈出不會跟隨按鈕，它會在其原始位置打開。

我用這個代碼：

(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];  
} 

你有任何意見或建議，以解決這個問題？ 謝謝！

Answer 1

嘿，嘗試這樣的事情：

- (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]; 
} 

我只是需要exactely一樣...代碼工作爲我在一個UIScrollView的UIButton。