iOS：弹出窗口位置

Question

我有这样的代码：

if([self.popoverController isPopoverVisible])
{
    [self.popoverController dismissPopoverAnimated:YES];
    return;
}

UIViewController* popoverContent = [[UIViewController alloc]init];

popoverContent.view = tableView1.view; 

self.popoverController = [[UIPopoverController alloc]
                          initWithContentViewController:popoverContent];

[self.popoverController presentPopoverFromRect:CGRectMake(1200, 280, 50, 50) inView:self.view permittedArrowDirections:UIPopoverArrowDirectionDown animated:TRUE];
[[[popoverController contentViewController] view] setAlpha:1]; //alpha del contenuto del popover

[popoverContent release];

它工作正常，但我无法将弹出窗口放在右下角，我尝试更改中的值 [self.popoverController PresentPopoverFromRect:CGRectMake(1200, 550, 100, 100) inView:self.view allowedArrowDirections:UIPopoverArrowDirectionDown 动画:TRUE];

但我想要这个弹出窗口在右下角很小......我该怎么办？

Answer 1

我是这样解决的：

[popoverController setPopoverContentSize:CGSizeMake(320.0f, 262.0f)];
[self.popoverController presentPopoverFromBarButtonItem:barButton permittedArrowDirections:UIPopoverArrowDirectionAny animated:YES];