xcode IOS：弹出框位置
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IOS: Popover position
提问by cyclingIsBetter
I have this code:
我有这个代码：
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];
It work fine, but I'm not able to place the popover in the lower right, I try to change values in [self.popoverController presentPopoverFromRect:CGRectMake(1200, 550, 100, 100) inView:self.view permittedArrowDirections:UIPopoverArrowDirectionDown animated:TRUE];
它工作正常，但我无法将弹出框放置在右下方，我尝试更改 [self.popoverController presentPopoverFromRect:CGRectMake(1200, 550, 100, 100) inView:self.view allowedArrowDirections:UIPopoverArrowDirectionDown 动画中的值：真的];
but I want this popover small in lower right...how can I do?
但我希望这个弹出框在右下角很小……我该怎么办？
采纳答案by cyclingIsBetter
I solved in this way:
我是这样解决的：
[popoverController setPopoverContentSize:CGSizeMake(320.0f, 262.0f)];
[self.popoverController presentPopoverFromBarButtonItem:barButton permittedArrowDirections:UIPopoverArrowDirectionAny animated:YES];
