警告:尝试在已经呈现 ViewController 的 ViewController 上呈现 ViewController

Warning: Attempt to present ViewController on ViewController which is already presenting ViewController

我有一个视图控制器,其工具栏带有 3 个 UIButton,可将新的视图控制器作为弹出窗口打开。我在 Storyboard 中创建了转场并选择了 "Present as Popover"。弹出窗口有效,但是当用户在当前打开弹出窗口的情况下点击另一个按钮时,我收到此错误:

Warning: Attempt to present <Fingerpainter.OpacityViewController: 0x79095110>  on <Fingerpainter.DrawingViewController: 0x7b278000> which is already presenting <Fingerpainter.BrushSizeViewController: 0x79573770>

有没有一种方法可以确保在打开新弹出窗口之前关闭所有弹出窗口?这是我在主要 ViewController(包含工具栏)中的 prepareForSegue 方法:

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
    let identifier = segue.identifier ?? ""
    let popoverPresentationController = segue.destinationViewController.popoverPresentationController
    popoverPresentationController!.delegate = self
switch identifier {
    case Storyboard.BrushSizeSegueIdentifier:
        if let brushSizeViewController = popoverPresentationController?.presentedViewController as? BrushSizeViewController {

            // set properties in brushSizeViewController
        }
    case Storyboard.OpacitySegueIdentifier:
        if let opacityViewController = popoverPresentationController?.presentedViewController as? OpacityViewController {
            //set properties in opacityViewController
        }
    case Storyboard.ColorSegueIdentity:
        if let colorViewController = popoverPresentationController?.presentedViewController as? ColorViewController {
            //set properties in colorViewController
        }
    default:
        break
    }

}

Is there a way to like make sure all popovers are closed before opening a new one

恰恰相反。 的工作是确保当弹出窗口存在时,召唤另一个弹出窗口的按钮不可点击。您可以通过禁用按钮来做到这一点,但更常见的是,为了协调按钮的禁用与弹出框的存在,这是通过调整弹出框呈现控制器的 passthroughViews.

来完成的

不幸的是,存在一个长期存在的巨大错误,即使将 passthroughViews 设置为 nil 也无法阻止工具栏按钮的点击。解决方法是延迟执行此操作。我的 popover 代码 lot 添加了这样的东西:

if let pop = popoverPresentationController {
    delay(0.1) {
        pop.passthroughViews = nil
    }
}

(此处描述 delay:)。