如何在iOS Swift 3中以编程方式推送和呈现给UIViewController而不进行选择

Question

我正在使用此代码在 iOS* Objective C中 以 编程方式推送 SHOW 和
MODALLY
。 现在想了解Swift 3。 *

NewsDetailsViewController *vc =  instantiateViewControllerWithIdentifier:@"NewsDetailsVCID"];
vc.newsObj = newsObj;
//--this(SHOW)
[self.navigationController pushViewController:vc animated:YES];  
//-- or this(MODAL)
[self presentViewController:vc animated:YES completion:nil];

Answer 1

推

喜欢

let storyboard = UIStoryboard(name: "Main", bundle: nil)
let vc = storyboard.instantiateViewControllerWithIdentifier("NewsDetailsVCID") as NewsDetailsViewController 
 vc.newsObj = newsObj
 navigationController?.pushViewController(vc,
 animated: true)

或更安全

  if let viewController = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "NewsDetailsVCID") as? NewsDetailsViewController {
        viewController.newsObj = newsObj
        if let navigator = navigationController {
            navigator.pushViewController(viewController, animated: true)
        }
    }

当下

   let storyboard = UIStoryboard(name: "Main", bundle: nil)
   let vc = self.storyboard?.instantiateViewControllerWithIdentifier("NewsDetailsVCID") as! NewsDetailsViewController
      vc.newsObj = newsObj
           present(vc!, animated: true, completion: nil)

或更安全

   if let vc = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "NewsDetailsVCID") as? NewsDetailsViewController
     {

     vc.newsObj = newsObj
    present(vc, animated: true, completion: nil)
    }