FPPopover不提供popover View iOS 6

时间:2012-11-27 09:06:30

标签: iphone ios ios5 ios6 popover

我正在尝试使用FPPopover打开通知视图,它可以在应用程序的iOS 5上运行,但不会在iOS 6上显示。但是,如果我使用相同的代码执行其他项目,它确实适用于iOS 5和6

这就是我在两个项目中所做的。

- (IBAction)NotificationButtonPressed:(id)sender {
    [self openNotificationsPopOver:sender];
}

- (void)openNotificationsPopOver:(id)sender {
    [self setNotificationCount:0];

    NotificationViewController *vc = [[UIStoryboard storyboardWithName:@"Notification" bundle:[NSBundle mainBundle]] instantiateViewControllerWithIdentifier:@"NotificationViewController"];

//    NotificationViewController *vc = [[NotificationViewController alloc]init];
//    [vc setParentViewController:self];


    FPPopoverController *popover = [[FPPopoverController alloc] initWithViewController:vc];

    //sender is the UIButton view
    [popover presentPopoverFromView:sender];

    //popover.arrowDirection = FPPopoverArrowDirectionAny;
    popover.tint = FPPopoverDefaultTint;

    popover.contentSize = CGSizeMake(200, 250);

    popover.arrowDirection = FPPopoverArrowDirectionAny;
    NSLog(@"Present popover");
    //sender is the UIButton view
    [popover presentPopoverFromView:sender];
//    [popover presentPopoverFromPoint:CGPointMake(160, 0)];

//sender is the UIButton view
//    [popover presentPopoverFromView:self.navigationController.navigationBar];
}

我尝试使用为其注释掉的行...所有方法都在单独的项目(iOS 5和6)上工作,但在主项目上它只适用于iOS 5

还尝试删除框架并再次添加它们。使用xCode 4.3启动了这个项目(如果重要的话,不知道)。

参考:https://github.com/50pixels/FPPopover

1 个答案:

答案 0 :(得分:0)

您需要确保在分配之前实例化“窗口”并显示弹出窗口。