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我已經創建了彈出窗口,其中有兩個按鈕。點擊彈出窗口中的按鈕時,我想關閉彈出窗口。如何在點擊按鈕時快速關閉彈出窗口?
這是我的代碼: FirstViewController:
@IBAction func bar_button(_ sender: UIBarButtonItem) {
let vc = storyboard?.instantiateViewController(withIdentifier: "SecondViewController") as!
SecondViewController
vc.preferredContentSize = CGSize(width: 200,height:80)
let navController = UINavigationController(rootViewController: vc)
navController.modalPresentationStyle = UIModalPresentationStyle.popover
let popover = navController.popoverPresentationController
popover?.delegate = self
popover?.barButtonItem = sender as! UIBarButtonItem
self.present(navController, animated: true, completion: nil)
}
SecondViewController:
@IBAction func second_button(_ sender: UIButton) {
//want to dismiss popover when button clicked
}
@IBAction func second_button(_ sender: UIButton) {
//want to dismiss popover when button clicked
}
但之後,它給出了這個警告 - 要求UIPopoverBackgroundVisualEffectView動畫的不透明度。這將導致該效果顯示中斷,直到不透明度返回到1. – Abhishek
@Abhishek:此問題已在下面的答案中簡要解決,請看看並讓我知道它是否解決您的問題,https://stackoverflow.com /問題/ 41589946/uipopoverbackgroundvisualeffectview,是幸福,問到有生命的,它的不透明度 – Bharath