This question already has answers here:
Share PDF through WhatsApp
(2 answers)
Closed 1 year ago.
I am using below code to send a text to a new WhatsApp number using a WhatsApp Url scheme instead of UIActivityViewController.
let urlWhats = "whatsapp://send?phone=+9197******2&text=Hi Hello"
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters:
CharacterSet.urlQueryAllowed){
if let whatsappURL = URL(string: urlString) {
if UIApplication.shared.canOpenURL(whatsappURL){
if #available(iOS 10.0, *) {
UIApplication.shared.open(whatsappURL, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(whatsappURL)
}
} else {
print("Install Whatsapp")
}
My current requirement is to share a Pdf file to a new WhatsApp number using WhatsApp Url scheme. Can anyone help me with this..
You can not pass the pdf file via url, it's make nonsense.
There are another way you can do is custom the UIActivityViewController.
You can refer this tutorial for UIActivityViewController customization: https://nshipster.com/uiactivityviewcontroller/
Related
// 1
let urlWhats = "https://wa.me/\(mobile)/?text=\(text)"
// 2
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed) {
// 3
if let whatsappURL = NSURL(string: urlString) {
// 4
if UIApplication.shared.canOpenURL(whatsappURL as URL) {
// 5
UIApplication.shared.open(whatsappURL as URL, options: [:], completionHandler: nil)
// UIApplication.shared.\
} else {
// 6
print("Cannot Open Whatsapp")
}
}
}
I'm able to launch whatsapp from my app from the above mentioned code, it is composing prefix text to the contact I wish to send and I need to click the send button in whatsapp manually . But I'm looking for a code which automatically sends whatsapp text to number from my app. Can anyone share your thoughts on this?
You can only compose the message for a particular contact using the Deep Linking method that you have used for it. For sending the message user has to click on the send button manually. You could provide the user with an alert that says so. But, it's not possible to do it for the user from your side. If you were able to send a message on Whatsapp by writing code without the user's confirmation it would be a break of user's privacy. Don't you think?
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)"
This question already has answers here:
Facebook not open page in app swift 3
(2 answers)
Closed 3 years ago.
I'm trying to open a facebook page from my app. Here is the code:
let facebookWebURL = URL.init(string: "https://www.facebook.com/aviramnet")
let facebookAppURL = URL.init(string: "fb://profile/220780122049926")
if (UIApplication.shared.canOpenURL(facebookAppURL!)){
UIApplication.shared.open(facebookAppURL!, options: [:], completionHandler: nil)
}else{
UIApplication.shared.open(facebookWebURL!, options: [:], completionHandler: nil)
}
But it only opens the app on Safari browser. Did I miss something?
And yes, I have the Facebook app installed.
you need to add in the app's info.plist :
under the LSApplicationQueriesSchemes array add an item:
fb for facebook, and instagram for instagram :
I am trying to share an audio file from my ios app to whatsapp, but with a direct opening of whatsapp, and not an opening of the sharing menu with all the tiles.
Here is what i have now:
// Getting the original file
let fileName = #MY FILE NAME#
let filePath = Bundle.main.path(forResource: fileName, ofType: "mp3")!
let urlData = URL.init(fileURLWithPath: filePath)
let nsData = NSData(contentsOf: urlData)
if (nsData != nil){
// Creating the temporary file to share in the accessible ressources
let newFileName = "file.mp3"
let newFilePath = "\(NSSearchPathForDirectoriesInDomains(.documentDirectory, .userDomainMask, true)[0])/\(newFileName)"
nsData?.write(toFile: newFilePath, atomically: true)
let newUrlData = URL.init(fileURLWithPath: newFilePath)
// Sharing the file to whatsapp
// Possibility 1 (does not work yet)
// let documentController = UIDocumentInteractionController(url: newUrlData)
// documentController.uti = "net.whatsapp.audio"
// documentController.presentOpenInMenu(from: CGRect.zero, in: self.view, animated: true)
// Possibility 2 (works only with the sharing menu)
let activityVC = UIActivityViewController(activityItems: [NSURL(fileURLWithPath: newFilePath)], applicationActivities: nil)
self.present(activityVC, animated: true, completion: nil)
}
As I do this, sharing an audio file to whatsapp works, but it first open the sharing menu, with the messenger tile, message tile, notes tile, ... (and it doesn't works for the messenger app). In the end I would like to be able to share on messenger AND whatsapp.
As explicated here in the whatsapp documentation, I want to open directly the whatsapp application when I try to share the file:
Alternatively, if you want to show only WhatsApp in the application list (instead of WhatsApp plus any other public/*-conforming apps) you can specify a file of one of aforementioned types saved with the extension that is exclusive to WhatsApp:
images - «.wai» which is of type net.whatsapp.image
videos - «.wam» which is of type net.whatsapp.movie
audio files - «.waa» which is of type net.whatsapp.audio
When triggered, WhatsApp will immediately present the user with the contact/group picker screen. The media will be automatically sent to a selected contact/group.
So I tried to change the line :
let newFileName = "file.mp3"
To one of these :
let newFileName = "file.mp3.waa"
let newFileName = "file.waa"
let newFileName = "file.waa.mp3"
But it still shows the same sharing menu (and can't read the audiofile if it ends with the .waa extension).
-> 1) Is it possible to do what I want to do ?
-> 2) If not, is there a way to share to messenger & whatsapp with the same code keeping one sharing menu
-> 3) If not, is there a way to reduce the sharing menu to only one tile depending on different calling event, so there is no confusing choosing of tiles
Thanks,
Antoine
Cf: XCODE / IOS - How to use exclusive extension to immediately present whatsapp (.wai, .waa, .wam)
FYI: As I went through a lot of tests with this, I couldn't find any solution yet.
Whatsapp recognize the file extension, but cannot even read it. Once shared, when you click on it, it's written ".whatsapp audio file", nothing more (And it's not even shared directly).
I sent a email to whatsapp developper team, they said they have others problem to fix currently, so it's not even on their to do list.
Wait & see..
I'm trying to open whatsapp conversation/chat for particular contact.
Instead of opening the desired chat it only opens the app.
No idea whats wrong :
let URLString = "whatsapp://send?abid=\(ID);text=lOL;"
UIApplication.sharedApplication().openURL(NSURL(string: URLString)!)
URLString value : whatsapp://send?abid=414;text=lOL
Any suggestions?
Update your URL like this:
whatsapp://send?abid=\(ID)&text=lOL
Source from HERE.
Try this and check if the UIApplication and open the URL.
let whatsAppURL: NSURL = NSURL(string: "whatsapp://send?abid=\(ID)&text=lOL")
if UIApplication.sharedApplication().canOpenURL(whatsAppURL){
UIApplication.sharedApplication().openURL(whatsAppURL)
}