How to open WIFI setting in Swift 3

Question

I want to open WIFI setting section from my iOS application, my code was working well before Swift 3 with iOS 9.2

if let settingsURL = URL(string: AppSetting         


        

Answer 1

Edited 26-02-20, Pay attention, in versions later than ios 10 your binary will be rejected

It's possible on iOS 10 and Swift 3

let url = URL(string: "App-Prefs:root=WIFI") //for WIFI setting app
let app = UIApplication.shared
app.openURL(url!)

Answer 2

We can't do this anymore on iOS 11, we can just open the settings :

if let url = URL(string:UIApplicationOpenSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
       let url =  UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

Answer 3

iOS 9+

You can't directly open the Wi-Fi settings tab from your app. You are just allowed to open the settings app in general.

The following code works with Swift 4 + iOS 9+:

func openWifiSettings() {
    print("Opening Wi-Fi settings.")

    let shared = UIApplication.shared
    let url = URL(string: UIApplicationOpenSettingsURLString)!

    if #available(iOS 10.0, *) {
        shared.open(url)
    } else {
        shared.openURL(url)
    }
}

Source: Apple developer documentation open and openURL

---

It was actually possible in iOS 9 to open the Wi-Fi settings tab directly (using private API URL schemes) but this was considered as bug usage.

Therefore: Don't use App-Prefs:root or pref:root URL schemes as they will lead to rejection of your app by Apple's review check.

Source: Apple developer forum post from Apple's eskimo.

---

iOS 11+

If you need to connect to a certain Wi-Fi network NEHotspotConfigurationManager maybe helps.

Source: Apple technical Q&A QA1942

Answer 4

Just use:

UIApplication.shared.openURL(URL(string:"prefs:root=WIFI")!)

Answer 5

 let url=URL(string: "App-Prefs:root=Privacy&path=Location")
        if UIApplication.shared.canOpenURL(url!)
        {
            UIApplication.shared.open(url!, options: [:], completionHandler: {sucess in

            })

        }
        else{
            UIApplication.shared.open(url!, options: [:], completionHandler: {sucess in

            })
        }

Answer 6

Use the following for iOS 10 and above:

if let url = URL(string:"App-Prefs:root=Settings&path=General") { 
    UIApplication.shared.openURL(url)
}