I have a cell in the settings table view where i want to send the users straight to the Appstore and open the app page so they can leave a review. There is plenty of content about that in Stackoverflow but none worked for me in iOS8 with Swift.
Thanks.
if let checkURL = NSURL(string: "http://www.itunes.com/yourAppLlink.html") {
if UIApplication.sharedApplication().openURL(checkURL) {
println("url sucefully opened")
}
} else {
println("invalid url")
}
Try this-
var path: NSURL = NSURL(string: "itms-apps://itunes.apple.com/us/app/YOUR_App_id")!
UIApplication.sharedApplication().openURL(path)
Related
We would really appreciate any help on the following. Through our app, the user can initiate a WhatsApp message (what happens is that the WhatsApp client starts with the phone + text preloaded, so the user just needs to tap the "send" button from the WhatsApp application).
We have an Android and an iOS app. In Android we are using the following code to select between WhatsApp and WhatsApp Business.
String url = "https://api.whatsapp.com/send?phone=" + phoneNumberToUse + "&text=" +
URLEncoder.encode(messageText, "UTF-8");
if(useWhatsAppBusiness){
intent.setPackage("com.whatsapp.w4b");
} else {
intent.setPackage("com.whatsapp");
}
URLEncoder.encode(messageText, "UTF-8");
intent.setPackage("com.whatsapp");
intent.setData(Uri.parse(url));
if (intent.resolveActivity(packageManager) != null) {
startActivity(intent);
} else {
Toast.makeText(this, "WhatsApp application not found", Toast.LENGTH_SHORT).show();
}
We are trying to achieve the same functionality on Swift for iOS, however, we did not find any way to programmatically define whether the OS should start WhatsApp or WhatsApp Business. The code listed below, always starts the one or other depending on which is installed. If both are installed, it starts the WhatsApp application.
let whatsApp = "https://wa.me/\(phoneNumber)/?text=\(shareableMessageText)"
guard let url = URL(string: whatsApp) else {
return
}
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: convertToUIApplicationOpenExternalURLOptionsKeyDictionary([:]), completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
So in simple words, is there any way, from our app, to select which WhatsApp application (WhatsApp or WhatsApp Business) is going to be launched?
Thanks
I have made some apps with WhatsApp but I had to use the web platform, not the business app.
You can check what app is installed in the device like this:
let app = UIApplication.shared
let appScheme = "App1://app"
if app.canOpenURL(URL(string: appScheme)!) {
print("App is install and can be opened")
} else {
print("App in not installed. Go to AppStore")
}
The 'App1' value must be changed for the app you want to check. WhatsApp App should use 'WhatsApp', and WhatsApp Business should use 'WhatsApp-Business'.
After that you can call the URL for each app, I mean, for WhatsApp you can use the URL with this format:
let whatsApp = "https://wa.me/\(phoneNumber)/?text=\(shareableMessageText)"
And for WhatsApp Business you have to use this format:
let whatsApp = "https://api.whatsapp.com/send?phone=\(phoneNumber)&text=\(shareableMessageText)"
It is possible too, that the first step was not necessary to do. Because of the call is made with the api, the device should open the Business app, and if it is made with the wa.me scheme, the device should open the WhatsApp as normal.
I am going to check my app to see if it is working or not.
UPDATE:
I have installed WhatsApp Business and I have made some test, with two different url calls.
The code I use is this:
let phoneNumber = "my_phone_number"
let shareableMessageText = "This_is_a_test_message"
let whatsApp = "https://wa.me/\(phoneNumber)/?text=\(shareableMessageText)"
if let urlString = whatsApp.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) {
if let whatsappURL = NSURL(string: urlString) {
if UIApplication.shared.canOpenURL(whatsappURL as URL) {
UIApplication.shared.openURL(whatsappURL as URL)
} else {
print("error")
}
}
}
Using this code you will see a prompt with a message giving you the option of send a message in WhatsApp or WhatsApp Business.
But if you use this other code:
let phoneNumber = "my_phone_number"
let shareableMessageText = "This_is_a_test_message"
let whatsApp = "whatsapp://send?phone=\(phoneNumber)&text=\(shareableMessageText)"
if let urlString = whatsApp.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) {
if let whatsappURL = NSURL(string: urlString) {
if UIApplication.shared.canOpenURL(whatsappURL as URL) {
UIApplication.shared.openURL(whatsappURL as URL)
} else {
print("error")
}
}
}
You will see a prompt asking you to open WhatsApp business. So the way to choose between WhatsApp and WhatsApp Business is the URL format. If you choose this format you will be ask to choose between one or another WA version:
let whatsApp = "https://wa.me/\(phoneNumber)/?text=\(shareableMessageText)"
But if you use this URL format, you will use WA Business directly:
let whatsApp = "whatsapp://send?phone=\(phoneNumber)&text=\(shareableMessageText)"
I am using universal link in iOS application. For the universal link I am using of Firebase dynamic link.
Universal link is working fine when app is installed. But when app is not installed then this link is opening in safari.
I am creating dynamic link (Universal link) programatically in Swift.
func createDynamicLinkForConsumer(){
guard let link = URL(string: "https://<Domain-Name>?consumerID=\(Auth.auth().currentUser?.uid ?? "")&type=consumerReferral") else { return }
let dynamicLinksDomainURIPrefix = "<appName>.page.link"
let linkBuilder = DynamicLinkComponents(link: link, domain: dynamicLinksDomainURIPrefix)
linkBuilder.options?.pathLength = .short
linkBuilder.iOSParameters = DynamicLinkIOSParameters(bundleID: "<Bundle-ID>")
linkBuilder.iOSParameters?.appStoreID = "<app-store-id>"
linkBuilder.androidParameters = DynamicLinkAndroidParameters(packageName: "<android package name>")
linkBuilder.navigationInfoParameters?.isForcedRedirectEnabled = true
guard let longDynamicLink = linkBuilder.url else { return }
print("The long URL is: \(longDynamicLink)")
self.referralLinkLabel.text = String(describing: longDynamicLink)
DynamicLinkComponents.shortenURL(longDynamicLink, options: nil) { (url, warnings, error) in
if url != nil {
print("Short URL is: \(url)")
self.referralLinkLabel.text = String(describing: url!)
}
}
}
This is the screen which opens after when safari opens the link:
After tapping on "Open" button of alert box browser opens app store app.
Actually what I would like to do is to redirect and go to the app store, so users can download the app directly. I don't want to this two steps to open app store.
You should first init DynamicLinkNavigationInfoParameters, then set property to true.
linkBuilder.navigationInfoParameters = DynamicLinkNavigationInfoParameters()
linkBuilder.navigationInfoParameters?.isForcedRedirectEnabled = true
Hi you can reduce these steps with this check on the dynamic link
But you should have in mind that you can have a problem with your link in Messanger app and some other apps.
if you click for example on someone’s Instagram button from my app
you gonna open the link instagram://app/ username in the profile
let’s say a user put “mark20” as Instagram username
You open the link instagram://app/mark20/
i want to open instagram with showing 'mark20' . currently it is open Instagram news feeds but not open 'mark20' profile . i want to button click open Instagram user profile page
For example
How can i do that
func shareToInstagram() {
//let instagramURL = URL(string: "instagram://app/")
//not working anymore
let instagramURL = URL(string: "instagram://app/mark20/")
if (UIApplication.shared.canOpenURL(instagramURL! as URL)) {
UIApplication.shared.open(instagramURL!, options: ["":""], completionHandler: nil)
} else {
print(" Instagram isn't installed ")
}
}
Yea I figure out the problem path parameter not working anymore but query parameter working
func shareToInstagram() {
let instagramURL = URL(string: "instagram://user?username=mark20")
if (UIApplication.shared.canOpenURL(instagramURL! as URL)) {
UIApplication.shared.open(instagramURL!, options: ["":""], completionHandler: nil)
print("miss you so much ")
} else {
print(" Instagram isn't installed ")
}
}
more check this link instagram iPhone Hooks
Swift
var instagramAppURL = URL(string: "instagram://user?username=USERNAME")
if UIApplication.shared.canOpenURL(instagramAppURL!) {
UIApplication.shared.openURL(instagramAppURL!)
}
Obj C
NSURL *instagramAppURL = [NSURL URLWithString:#"instagram://user?username=USERNAME"];
if ([[UIApplication sharedApplication] canOpenURL:instagramAppURL]) {
[[UIApplication sharedApplication] openURL:instagramAppURL];
}
First, you have to modify Info.plist to list instagram and facebook with LSApplicationQueriesSchemes. Simply open Info.plist as a Source Code, and paste this:
<key>LSApplicationQueriesSchemes</key>
<array>
<string>instagram</string>
<string>fb</string>
</array>
For more details and more ways please refer below link
https://www.instagram.com/developer/mobile-sharing/iphone-hooks/
I know there are lot of topics about this problem but is there a solution? I have not found official documentation for that.
My problem is I need to redirect my code on storage settings like this : (work in iOS 9)
let settingsUrl = NSURL(string: "prefs:root=CASTLE&path=STORAGE_AND_BACKUP")
if let url = settingsUrl {
UIApplication.shared.openURL(url as URL)
}
But since ios10 this method don't work, so Is there an alternative? I saw SnapChat, Google Maps redirect their apps to different part on settings (not the main screen of settings) so I think there is a solution.
I already implement the url scheme in info.plist but it's still not work
I tried this method but same issue too
let settingsUrl = NSURL(string: "prefs:root=CASTLE&path=STORAGE_AND_BACKUP")
if let url = settingsUrl {
if #available(iOS 10.0, *) {
UIApplication.shared.open(URL.init(string:"Prefs:root=General&path=STORAGE_ICLOUD_USAGE/DEVICE_STORAGE")!, options: [UIApplicationOpenURLOptionUniversalLinksOnly:true], completionHandler:{(success: Bool?) -> Void in})
} else {
UIApplication.shared.openURL(url as URL)
}
}
Thank in advance.
Since iOS 10, it's not possible to open the Settings app from a third party app. The only settings that are allowed to be opened are Keyboard setting but only by a custom keyboard extension and your own application settings.
More details: here
Note: Even for iOS 9, using the URL string that is mentioned in the question can lead to app rejection as it violates iOS App Review Guidelines.
So I am about to launch an app to the App Store. My issue is that I have a rate my app please button but I do not know the right code to insert there.
My fiends tried this on their app and said it was no good:
Does anybody know how I can fix this issue?
let iTunesReviewPageLink = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?id=1073785561&pageNumber=0&sortOrdering=2&type=Purple+Software&mt=8"
// Go directly to review page on App Store
if let url = NSURL(string: iTunesReviewPageLink) {
UIApplication.sharedApplication().openURL(url)
}
The only unknown thing is the ID, right? You can see the ID of your app before it is published - once you set it up in iTunes Connect.
If your app is not yet released, you haven't got an App Store link. So that's impossible.
In order to implement this functionality when your app is released, you can use the following code:
Swift 2
let appIDString = "APP_ID" // your app ID
let reviewsURLString = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?pageNumber=0&sortOrdering=1&type=Purple+Software&mt=8&id=\(appIDString)"
let reviewsURL = NSURL(string: reviewsURLString)
if reviewsURL != nil && UIApplication.sharedApplication().canOpenURL(reviewsURL!) {
UIApplication.sharedApplication().openURL(reviewsURL!)
}
else {
// handle situation if reviews url cannot be opened.
}
Swift 3
let appIDString = "APP_ID" // your app ID
let reviewsURLString = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?pageNumber=0&sortOrdering=1&type=Purple+Software&mt=8&id=\(appIDString)"
let reviewsURL = URL(string: reviewsURLString)
if reviewsURL != nil && UIApplication.shared.canOpenURL(reviewsURL!) {
UIApplication.shared.openURL(reviewsURL!)
}
else {
// handle situation if reviews url cannot be opened.
}
EDIT:
This links works in iOS 8 and 9 and links directly to reviews page of the app in App Store application.
I am not sure about iOS 7. Probably for iOS 7 you need to use different link.