如何仅用一个按钮快速消除弹出窗口 [英] How to dismiss a popover only with a button in swift

问题描述

我的iPad应用程序具有多个数据收集弹出窗口，并且我希望能够通过触摸它的外部来禁用弹出窗口的关闭，然后我将使用一个按钮，由用户自行决定退出该弹出窗口.

My iPad app has several data gathering popovers, and I want to be able to disable the dismissal of the popover by touching outside of it, I then will use a button to quit the popover at the users discretion.

该应用程序看起来不错，弹出窗口工作正常，并且其中有一个按钮可以很好地退出.只有我找不到在Swift中禁用解雇的方法，在obj-c上有很多帖子，但是在Swift中什么都没有.

The app looks great, the popovers work fine, and I have a button inside them that quits nicely. Only I can't find a way of disabling dismissal in Swift, lots of posts on obj-c but nothing in Swift.

这是否意味着该功能不再可用?

Does this mean that the functionality is no longer available?

非常感谢您为我的挫败感提供帮助.

I would greatly appreciate any help to my frustration.

推荐答案

只需将视图控制器的modalInPopover设置为true，将弹出窗口的passthroughViews设置为nil.但是您必须使用延迟的性能来执行后者，否则将无法正常工作.只需稍作延迟.示例:

Simply set the view controller's modalInPopover to true and the popover's passthroughViews to nil. But you must do the latter using delayed performance or it won't work. A small delay is all that's needed. Example:

    let vc = UIViewController()
    vc.modalPresentationStyle = .Popover
    self.presentViewController(vc, animated: true, completion: nil)
    if let pop = vc.popoverPresentationController {
        vc.modalInPopover = true
        delay(0.1) {
            pop.passthroughViews = nil
        }
    }

有关delay功能，请参见 dispatch_after-快速运行GCD吗?.

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