Question

我在该项目中的一个Demo项目中创建了可达性我有两个视图控制器，如果有互联网连接我正在显示OnlineViewController如果没有互联网连接我正在显示OfflineViewController。请参阅我在Appdelegate.swift中实现的以下代码。

 var reachability:Reachability?
    func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {

        NotificationCenter.default.addObserver(self, selector: #selector(reachabilityStatusChanged(_:)), name: .reachabilityChanged, object: nil)
        reachability = Reachability.forInternetConnection()
        reachability!.startNotifier()
            return true
    }

我的功能就像这个

 @objc func reachabilityStatusChanged(_ sender: NSNotification) {

        var remoteHostStatus = self.reachability!.currentReachabilityStatus()

        if (remoteHostStatus == NotReachable)
        {
            print ("no net")
            let testController = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "OfflineViewController") as! OfflineViewController
            window!.rootViewController = testController
            window!.makeKeyAndVisible()
        }
        else
        {
            let testController = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "OnlineViewController") as! OnlineViewController
            window!.rootViewController = testController
            window!.makeKeyAndVisible()
            print (" wifi")

        }

    }

现在我的问题是我想在该项目的实时项目中实现相同的功能，我有3个故事板和多个视图控制器。那么如果有互联网连接，如何再次显示用户当前屏幕。

Answer 1

我用以下代码解决了我的问题。我不确定我的答案是对还是错。

@objc func reachabilityStatusChanged(_ sender: NSNotification) {

        var remoteHostStatus = self.reachability!.currentReachabilityStatus()

        if (remoteHostStatus == NotReachable)
        {
            print ("no net")
            let testController = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "OfflineViewController") as! OfflineViewController
            window!.rootViewController = testController
            window!.makeKeyAndVisible()
        }
        else
        {
          UserTypeValidateCall()

        }

    }.

//MARK:- Validating User Type
func  UserTypeValidateCall()
{
    let Usertype  =  UserDefaults.standard.string(forKey: "Usertype")
    print("Usertype-----",Usertype)

    if(( Usertype) != nil)
    {
        if (Usertype?.isEqual("Parent"))!
        {
            ParentDashBoardEnter()
        }
        else  if (Usertype?.isEqual("Employee"))!
        {
            TeacherDashBoardEnter()
        }
        else if (Usertype?.isEqual("Admin"))!
        {
            AdminDashBoardEnter()
        }
    }
    else
    {
        LoginScreenEnter()
    }
}

Answer 2

同样在项目中使用此函数并将用户导航到特定的视图控制器。

有关详细信息，请参阅此链接。 Check for internet connection with Swift

Answer 3

import UIKit
import Reachability
class CommonUtility: NSObject {

    class func isConnected() -> Bool{
        let reachability:Reachability = Reachability.forInternetConnection()
        let networkStatus = reachability.currentReachabilityStatus().rawValue
        return networkStatus != 0
   }
}

Answer 4

//使用以下链接下载它，只需将Reachability Folder拖放到您的项目

即可

https://github.com/ashleymills/Reachability.swift

或

pod'ReachabilitySwift'，'〜＆gt; 4.1'

注意：这里的ReachabilitySwift版本是基于你必须要的swift版本

let reachability = Reachability()!

  override func viewDidLoad()
  {
    super.viewDidLoad()
  }


 if (reachability.isReachable)
   {
         // your code
   }
 else
   {
       // your code
   }

面对可达性问题IOS Swift

4 个答案: