Sending message to WhatsApp from your app using Swift? - ios

For one of my app, I wanted to share data to WhatsApp contacts. I tried few solutions overs the StackOverflow but couldn't get exact solution. After some trials could achieve what I was looking for, so sharing here for anyone's future reference.

var url = NSURL(string: "whatsapp://send?text=Hello%20Friends%2C%20Sharing%20some%20data%20here...%20!")
//Text which will be shared on WhatsApp is: "Hello Friends, Sharing some data here... !"
if UIApplication.sharedApplication().canOpenURL(url!) {
UIApplication.sharedApplication().open(url as URL, options: [:]) { (success) in
if success {
print("WhatsApp accessed successfully")
} else {
print("Error accessing WhatsApp")
}
}
}
Note: text needs to be URL encoded. You can get it using any of the open source tools over internet or using addingPercentEncoding(withAllowedCharacters:) function in iOS.
e.g.
var urlString = "Hello Friends, Sharing some data here... !"
var urlStringEncoded = urlString.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed)
var url = NSURL(string: "whatsapp://send?text=\(urlStringEncoded!)")

Swift 5
Code
let urlWhats = "whatsapp://send?text=\("Hello World")"
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed) {
if let whatsappURL = NSURL(string: urlString) {
if UIApplication.shared.canOpenURL(whatsappURL as URL) {
UIApplication.shared.open(whatsappURL as URL)
}
else {
print("please install watsapp")
}
}
}

Swift 3.0
Try with this code for access watsapp in your application. Its working perfectly for me.
#IBAction func sendButtonAction(_ sender: Any)
{
let date = Date()
let msg = "Hi my dear friends\(date)"
let urlWhats = "whatsapp://send?text=\(msg)"
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed) {
if let whatsappURL = NSURL(string: urlString) {
if UIApplication.shared.canOpenURL(whatsappURL as URL) {
UIApplication.shared.openURL(whatsappURL as URL)
} else {
print("please install watsapp")
}
}
}
}

Addition to above solutions, starting from iOS 9, we need to add whatsapp to LSApplicationQueriesSchemes key in info.plist. After this only it worked for me.

My code is Looking Like this
let encodeQuizStr = "Check Out The Quiz With link \n http://www.proprofs.com "
let urlQuizStringEncoded = encodeQuizStr.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed)
guard let whatsAppUrl = NSURL(string: "whatsapp://send?text="+urlQuizStringEncoded!) else { return }
if UIApplication.shared.canOpenURL(whatsAppUrl as URL) {
if #available(iOS 10.0, *) {
print(urlQuizStringEncoded!)
UIApplication.shared.open(whatsAppUrl as URL, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(whatsAppUrl as URL)
}
}
else{
ProjectUtility.AlertWith(self, message: " What's App is Not Available.", Title: "Sorry")
}
working fine But When I put This URL
("http://www.proprofs.com/quiz-school/story.php?title=pq-find-out-which-ice-age-character-you-are ")
Then Its Not Working Please Check Thanks.HelpWill Be Appriciated.

Swift 5
Please follow the below steps for sharing on WhatsApp through URL Schemes
Add this code in your app "info.plist"
<key>LSApplicationQueriesSchemes</key>
<array>
<string>whatsapp</string>
</array>
For sharing text and URL
Code
#IBAction func whatsappShareText(_ sender: AnyObject) {
let message = "First Whatsapp Share & https://www.google.co.in"
var queryCharSet = NSCharacterSet.urlQueryAllowed
// if your text message contains special characters like **+ and &** then add this line
queryCharSet.remove(charactersIn: "+&")
if let escapedString = message.addingPercentEncoding(withAllowedCharacters: queryCharSet) {
if let whatsappURL = URL(string: "whatsapp://send?text=\(escapedString)") {
if UIApplication.shared.canOpenURL(whatsappURL) {
UIApplication.shared.open(whatsappURL, options: [: ], completionHandler: nil)
} else {
debugPrint("please install WhatsApp")
}
}
}
}
Happy Coding!

As per their FAQ, you should be using the universal links instead:
https://wa.me/1234567890
Reference: https://faq.whatsapp.com/563219570998715/?locale=en_US

Related

Send Text and Link to Snapchat from my Swift app without the SDK

I am integrating sharing options from my app to Snapchat.
I have a dynamic URL obtained in an object and clicking the Snapchat's share button directly opens the app if Snapchat is there on the device and show the text with the link. I am using the below code to share which gives an error on Snapchat. Below is my Code.
func shareTextOnSnapchat(obj:VideoData) {
let shaUrl = URL(string: obj.share_url ?? "")
if let myURL:URL = shaUrl{
let promoText = "Check out this great new video from \(obj.name ?? ""), I found on talent app"
let shareString = "snapchat://text=\(promoText)&url=\(myURL)"
let escapedShareString = shareString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlQueryAllowed)!
let url = URL(string: escapedShareString)
UIApplication.shared.openURL(url!)
}
}
I have used this to post video to snapchat. You have option to either post text or a video.
Pod used
pod 'SnapSDK', :subspecs => ['SCSDKCreativeKit']
import SCSDKCreativeKit
var scInstalled = false
override func viewDidLoad() {
super.viewDidLoad()
scInstalled = schemeAvailable(scheme: "snapchat://")
}
func ShowSnapchat(){
if scInstalled {
//shareTextOnSnapchat(obj:videoObj ?? VideoData())
shareFileOnSnapchat(obj:videoObj ?? VideoData())
}else{
downloadSharingAppAlert(appName:"Snapchat")
}
}
func shareTextOnSnapchat(obj:VideoData) {
let shaUrl = URL(string: obj.share_url ?? "")
if let myURL:URL = shaUrl{
let originalString = "\(myURL)"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters:CharacterSet.urlQueryAllowed)
//let url = URL(string: "snapchat://snap?text=\(escapedString!)")
let url = URL(string: "https://www.snapchat.com/send?text=\(escapedString!)")
if UIApplication.shared.canOpenURL(url! as URL)
{
UIApplication.shared.open(url! as URL, options: [:], completionHandler: nil)
}
}
}
func shareFileOnSnapchat(obj:VideoData){
//// SHARE VIDEO
LoadingOverlay.shared.showLoaderView(view: self.view)
let shaUrl = URL(string: obj.output_vid ?? "")
if let myURL:URL = shaUrl{
let snapVideo = SCSDKSnapVideo(videoUrl: myURL)
let snapContent = SCSDKVideoSnapContent(snapVideo: snapVideo)
// Send it over to Snapchat
snapAPI.startSending(snapContent) { (error) in
if let error = error {
print(error.localizedDescription)
LoadingOverlay.shared.hideLoaderView()
MyCustomAlert.sharedInstance.ShowAlert(vc: self, myTitle: "", myMessage: StringClass.sharedInstance.lcStr_oopsSomethingWentwrong)
} else {
// success
print("Posted to snapchat")
LoadingOverlay.shared.hideLoaderView()
MyCustomAlert.sharedInstance.ShowAlert(vc: self, myTitle: "", myMessage: StringClass.sharedInstance.lcStr_postedToSnapchat)
}
}
}
}
}
func downloadSharingAppAlert(appName:String){
var appStoreURL = "https://apps.apple.com/in/app/snapchat/id447188370"
//Open Appstore for Download
}

How to open WhatsApp from Swift app?

I am using webview for my Swift app and I have "Share on WhatsApp" button on my website which works fine on a browser. But on iPhone app, when I click on the button, nothing happens.
How to open WhatsApp from my app? I am using Xcode 8 and iOS 10.
For Swift 4.2+ and iOS 9+
Method 1: (launches WhatsApp application if is installed)
let phoneNumber = "+989160000000" // you need to change this number
let appURL = URL(string: "https://api.whatsapp.com/send?phone=\(phoneNumber)")!
if UIApplication.shared.canOpenURL(appURL) {
if #available(iOS 10.0, *) {
UIApplication.shared.open(appURL, options: [:], completionHandler: nil)
}
else {
UIApplication.shared.openURL(appURL)
}
}
Method 2:(open WhatsApp short-link web page using safari)
let phoneNumber = "+989160000000" // you need to change this number
let appURL = URL(string: "https://wa.me/\(phoneNumber)")!
if UIApplication.shared.canOpenURL(appURL) {
if #available(iOS 10.0, *) {
UIApplication.shared.open(appURL, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(appURL)
}
}
Note : '+' in phone number is OK.
I know this is an old question, but the following worked for me (I'm using xcode 8.3.3 and swift 3).
I added whatsapp query scheme inside Info.plist.
<key>LSApplicationQueriesSchemes</key>
<array>
<string>whatsapp</string>
</array>
Info.plist
After adding it, the following works:
let urlString = "whatsapp://send?text=Message to share"
let urlStringEncoded = urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
let URL = NSURL(string: urlStringEncoded!)
if UIApplication.shared.canOpenURL(URL! as URL) {
UIApplication.shared.openURL(URL! as URL)
}
UIApplication.shared.openURL(URL(string:"https://api.whatsapp.com/send?phone=phoneNumber")!)
phoneNumber might be with (+) or not.
phoneNumber looks like 99455555555 or +99455555555
Devs Here is my Code for Opening WhatsApp Chat in Xcode 13.0 and iOS 15.0 for specific Contact.
func navigateToWhatsApp() {
var countryCode = "91". //Country code
var mobileNumber = "1234567890" //Mobile number
let urlString = "https://api.whatsapp.com/send?phone=\(countryCode)\(mobileNumber)"
let urlStringEncoded = urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
let URL = NSURL(string: urlStringEncoded!)
if UIApplication.shared.canOpenURL(URL! as URL) {
debugPrint("opening Whatsapp")
UIApplication.shared.open(URL as! URL, options: [:]) { status in
debugPrint("Opened WhatsApp Chat")
}
} else {
debugPrint("Can't open")
}
}
For this you should use URL schemes.
let message = "Message"
let urlWhats = "whatsapp://send?text=\(message)"
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed) {
if let whatsappURL = NSURL(string: urlString) {
if UIApplication.shared.canOpenURL(whatsappURL as URL) {
UIApplication.shared.open(whatsappURL as URL, options: [:], completionHandler: { (Bool) in
})
} else {
// Handle a problem
}
}
}

How to open Google Maps to show route using Swift

I read around the Internet but couldn't find a legit answer. I need to open Google Maps when the user clicks a button to show directions. Start and destination must be automatically filled.
How can I achieve this in Swift?
If anyone has a solution, please provide me with a sample code. The start is always going to be the user's current location.
OK I found the answer myself.
If you want to show directions from the user's current location, leave the field saddr blank and in the field daddr you can enter the destination coordinates.
This is how I did it
if (UIApplication.sharedApplication().canOpenURL(NSURL(string:"comgooglemaps://")!)) {
UIApplication.sharedApplication().openURL(NSURL(string:
"comgooglemaps://?saddr=&daddr=\(place.latitude),\(place.longitude)&directionsmode=driving")!)
} else {
NSLog("Can't use comgooglemaps://");
}
}
for any further queries you can refer to this link Google Map URL Scheme
Swift 5 - Code
func openGoogleMap() {
guard let lat = booking?.booking?.pickup_lat, let latDouble = Double(lat) else {Toast.show(message: StringMessages.CurrentLocNotRight);return }
guard let long = booking?.booking?.pickup_long, let longDouble = Double(long) else {Toast.show(message: StringMessages.CurrentLocNotRight);return }
if (UIApplication.shared.canOpenURL(URL(string:"comgooglemaps://")!)) { //if phone has an app
if let url = URL(string: "comgooglemaps-x-callback://?saddr=&daddr=\(latDouble),\(longDouble)&directionsmode=driving") {
UIApplication.shared.open(url, options: [:])
}}
else {
//Open in browser
if let urlDestination = URL.init(string: "https://www.google.co.in/maps/dir/?saddr=&daddr=\(latDouble),\(longDouble)&directionsmode=driving") {
UIApplication.shared.open(urlDestination)
}
}
}
Don't forget to write this in info.plist
<key>LSApplicationQueriesSchemes</key>
<array>
<string>comgooglemaps</string>
<string>comgooglemaps-x-callback</string>
</array>
The Answer is already there but in older versions of Swift
In Swift 3
//Working in Swift new versions.
if (UIApplication.shared.canOpenURL(URL(string:"comgooglemaps://")!))
{
UIApplication.shared.openURL(NSURL(string:
"comgooglemaps://?saddr=&daddr=\(Float(latitude!)),\(Float(longitude!))&directionsmode=driving")! as URL)
} else
{
NSLog("Can't use com.google.maps://");
}
For those that the solution for the accepted answer didn't work for them, remember to add comgooglemaps to LSApplicationQueriesSchemes in the info.plist.
#IBAction func NavigationTrackerPressedPressed(_ sender: UIButton){
if let UrlNavigation = URL.init(string: "comgooglemaps://") {
if UIApplication.shared.canOpenURL(UrlNavigation){
if self.destinationLocation?.longitude != nil && self.destinationLocation?.latitude != nil {
let lat = (self.destinationLocation?.latitude)!
let longi = (self.destinationLocation?.longitude)!
if let urlDestination = URL.init(string: "comgooglemaps://?saddr=&daddr=\(lat),\(longi)&directionsmode=driving") {
UIApplication.shared.openURL(urlDestination)
}
}
}
else {
NSLog("Can't use comgooglemaps://");
self.openTrackerInBrowser()
}
}
else
{
NSLog("Can't use comgooglemaps://");
self.openTrackerInBrowser()
}
}
func openTrackerInBrowser(){
if self.destinationLocation?.longitude != nil && self.destinationLocation?.latitude != nil {
let lat = (self.destinationLocation?.latitude)!
let longi = (self.destinationLocation?.longitude)!
if let urlDestination = URL.init(string: "https://www.google.co.in/maps/dir/?saddr=&daddr=\(lat),\(longi)&directionsmode=driving") {
UIApplication.shared.openURL(urlDestination)
}
}
}
Here is the Swift 3 Update
First, you need to two items to LSApplicationQueriesSchemes in the info.plist.
Then you can use this function to load address on Google maps.
let primaryContactFullAddress = "No 28/A, Kadalana, Moratuwa, Sri Lanka"
#IBAction func showLocaionOnMaps(_ sender: Any) {
let testURL: NSURL = NSURL(string: "comgooglemaps-x-callback://")!
if UIApplication.shared.canOpenURL(testURL as URL) {
if let address = primaryContactFullAddress.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed) {
let directionsRequest: String = "comgooglemaps-x-callback://" + "?daddr=\(address)" + "&x-success=sourceapp://?resume=true&x-source=AirApp"
let directionsURL: NSURL = NSURL(string: directionsRequest)!
let application:UIApplication = UIApplication.shared
if (application.canOpenURL(directionsURL as URL)) {
application.open(directionsURL as URL, options: [:], completionHandler: nil)
}
}
} else {
NSLog("Can't use comgooglemaps-x-callback:// on this device.")
}
}
Swift 4 Working Code
if let url = URL(string: "comgooglemaps://?saddr=&daddr=\(location.coordinate.latitude),\(location.coordinate.longitude)&directionsmode=driving") {
UIApplication.shared.open(url, options: [:])
}
this code works for me. And ı didn't insert
* LSApplicationQueriesSchemes
If you use emulator can't see the results. Don't forget work your project on the Phone.
1) Method: You can pass the only destination address current address will be automatically fetched by google map application.
let strLat : String = "23.035007"
let strLong : String = "72.529324"
override func viewDidLoad() {
super.viewDidLoad()
if (UIApplication.shared.canOpenURL(URL(string:"comgooglemaps://")!)) {
UIApplication.shared.openURL(URL(string:"comgooglemaps://?saddr=&daddr=\(strLat),\(strLong)&directionsmode=driving")!)
}
else {
print("Can't use comgooglemaps://");
}
}
2) You can pass both start and destination address
let strLat : String = "23.035007"
let strLong : String = "72.529324"
let strLat1 : String = "23.033331"
let strLong2 : String = "72.524510"
override func viewDidLoad() {
super.viewDidLoad()
if (UIApplication.shared.canOpenURL(URL(string:"comgooglemaps://")!)) {
UIApplication.shared.openURL(URL(string:"comgooglemaps://?saddr=\(strLat),\(strLong)&daddr=\(strLat1),\(strLong2)&directionsmode=driving&zoom=14&views=traffic")!)
}
else {
print("Can't use comgooglemaps://");
}
}
I want the possibility that the user can open Google Maps via browser (only works when user hasn't the Google Maps App)
Google gives a documentation for that here.
Since iOS 9 you have to set a scheme. This you will find here.
For me the google solution doesn't work. Maybe you are smarter and find a solution (please post!).
Anyway I made a simple web call:
let lat = 37.7
let lon = -122.4
if (UIApplication.sharedApplication().canOpenURL(NSURL(string:"https://maps.google.com")!))
{
UIApplication.sharedApplication().openURL(NSURL(string:
"https://maps.google.com/?q=#\(lat),\(lon)")!)
}
This could be use as a kind of fallback from sumeshs answer.
let lat = self.upcomingListArr[indexPath.item].latitude!
let long = self.upcomingListArr[indexPath.item].longitude!
if (UIApplication.shared.canOpenURL(NSURL(string:"comgooglemaps://")! as URL)) {
UIApplication.shared.openURL(NSURL(string:
"comgooglemaps://?saddr=&daddr=\(String(describing: lat)),\(String(describing: long))")! as URL)
} else {
UIApplication.shared.openURL(NSURL(string:
"https://www.google.co.in/maps/dir/?saddr=&daddr=\(String(describing: lat)),\(String(describing: long))")! as URL)
}
You can create small builder for URL, that will redirect you to GoogleMaps
/*
add to InfoPlist:
<key>LSApplicationQueriesSchemes</key>
<array>
<string>comgooglemaps</string>
<string>comgooglemaps-x-callback</string>
</array>
*/
final public class GoogleMapsURIConstructor {
public enum NavigationType: String {
case driving
case transit
case walking
}
public class func prepareURIFor(latitude lat: Double,
longitude long: Double,
fromLatitude fromLat: Double? = nil,
fromLongitude fromLong: Double? = nil,
navigation navigateBy: NavigationType) -> URL? {
if let googleMapsRedirect = URL(string: "comgooglemaps://"),
UIApplication.shared.canOpenURL(googleMapsRedirect) {
if let fromLat = fromLat,
let fromLong = fromLong {
let urlDestination = URL(string: "comgooglemaps-x-callback://?saddr=\(fromLat),\(fromLong)?saddr=&daddr=\(lat),\(long)&directionsmode=\(navigateBy.rawValue)")
return urlDestination
} else {
let urlDestination = URL(string: "comgooglemaps-x-callback://?daddr=\(lat),\(long)&directionsmode=\(navigateBy.rawValue)")
return urlDestination
}
} else {
if let fromLat = fromLat,
let fromLong = fromLong {
let urlDestination = URL(string: "https://www.google.co.in/maps/dir/?saddr=\(fromLat),\(fromLong)&daddr=\(lat),\(long)&directionsmode=\(navigateBy.rawValue)")
return urlDestination
} else {
let urlDestination = URL(string: "https://www.google.co.in/maps/dir/?saddr=&daddr=\(lat),\(long)&directionsmode=\(navigateBy.rawValue)")
return urlDestination
}
}
}
}
usage:
let uri = GoogleMapsURIConstructor.prepareURIFor(latitude: lat,
longitude: long,
fromLatitude: userLatitude,
fromLongitude: userLongitude,
navigation: .driving)
UIApplication.shared.open(uri, options: [: ], completionHandler: nil)
result:
First off all, thank you guys for your answers beyond. But when I'm working around, I didn't find the working answer for me. And as result I've made working code with two cases: when user has the google maps app and if not.
Here is my version of code. Xcode 10, Swift 4.2. Hope it will help.
Step 1: Find your info.plist and Open As -> Source Code
Step 2: Add these lines between <dict> and </dict>
Step 3: In your action add this code. Don't forget to set your latitude and longitude.
let latitude = 44.987781
let longitude = 88.987781
let appDomen: String = "comgooglemaps://"
let browserDomen: String = "https://www.google.co.in/maps/dir/"
let directionBody: String = "?saddr=&daddr=\(latitude),\(longitude)&directionsmode=driving"
// Make route with google maps application
if let appUrl = URL(string: appDomen), UIApplication.shared.canOpenURL(appUrl) {
guard let appFullPathUrl = URL(string: appDomen + directionBody) else { return }
UIApplication.shared.openURL(appFullPathUrl)
// If user don't have an application make route in browser
} else if let browserUrl = URL(string: browserDomen), UIApplication.shared.canOpenURL(browserUrl) {
guard let browserFullPathUrl = URL(string: browserDomen + directionBody) else { return }
UIApplication.shared.openURL(browserFullPathUrl)
}
If google map app not installed, open with apple map version.
import Foundation
import CoreLocation
import UIKit
import MapKit
struct GoogleMapApp {
func openWithDirection(cordinate: CLLocationCoordinate2D, locationName: String) {
let hasGoogleMapApp = UIApplication.shared.canOpenURL(URL(string: "comgooglemaps://")!)
if hasGoogleMapApp {
let urlString = "comgooglemaps-x-callback://?saddr=&daddr=\(cordinate.latitude),\(cordinate.longitude)&directionsmode=driving"
if let url = URL(string: urlString) {
UIApplication.shared.open(url, options: [:])
}
} else {
// If google map not installed, open with Apple Map App
let mapItem = MKMapItem(placemark: MKPlacemark(coordinate: cordinate, addressDictionary: nil))
mapItem.name = locationName
mapItem.openInMaps(launchOptions: [MKLaunchOptionsDirectionsModeKey: MKLaunchOptionsDirectionsModeDriving])
}
}
}
// I hope it will work perfactly.
let googleURL = NSURL(string: "comgooglemaps://?q=")
if(UIApplication.shared.canOpenURL(googleURL! as URL)) {
UIApplication.shared.open(URL(string:"comgooglemaps://?saddr=\(DEVICE_LAT),\(DEVICE_LONG)&daddr=\(addressLat),\(addressLng)&directionsmode=driving")!, options: [:], completionHandler: nil)
}
//Also set permission from info.plist according to attached image.
enter image description here
We can use this method to open google maps. If google maps is not installed it will redirect to browser.
func redirectToGoogleMap(coordinates : String) {
if let url = URL(string: "comgooglemaps://?saddr=&daddr=\(coordinates)&directionsmode=driving"),
UIApplication.shared.canOpenURL(url) {
print("comgoogle maps worked")
if #available(iOS 10, *) {
UIApplication.shared.open(url, options: [:], completionHandler:nil)
} else {
UIApplication.shared.openURL(url)
}
} else{
if let url = URL(string: "https://www.google.co.in/maps/dir/?saddr=&daddr=\(coordinates)&directionsmode=driving"),
UIApplication.shared.canOpenURL(url) {
print("google com maps worked")
if #available(iOS 10, *) {
UIApplication.shared.open(url, options: [:], completionHandler:nil)
} else {
UIApplication.shared.openURL(url)
}
}else{
kSweetAlert.showAlert(CommonMethods.setLangText("oops"), subTitle: "Oops! Could not open google maps", style: AlertStyle.error)
}
}
}`
SWIFT 5.1 (WORKED FOR ME)
// Destination Address (Lat/Long)
let place = business?.businessAddresse
// Google URL
let stringURL = "comgooglemaps://"
// Checking Nil
if !(place?.latitude == nil) || !(place?.longitude == nil) {
if UIApplication.shared.canOpenURL(URL(string: stringURL)!) {
// If have Google Map App Installed
if let url = URL(string: "comgooglemaps://?saddr=&daddr=\(place?.latitude ?? 0.0),\(place?.longitude ?? 0.0)&directionsmode=driving") {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
} else {
// If have no Google Map App (Run Browser Instead)
if let destinationURL = URL(string: "https://www.google.co.in/maps/dir/?saddr=&daddr=\(place?.latitude ?? 0.0),\(place?.longitude ?? 0.0)&directionsmode=driving") {
UIApplication.shared.open(destinationURL, options: [:], completionHandler: nil)
}
}
} else {
// Show Alert When No Location Found
self.delegate?.noDirectionError(title: "Error", message: "There's no direction available for this restaurant")
}
Note: This code only work on iOS Version 11 or Later, I hope this helps you guys too, cheer.
Here is the simple solution to choose navigation application between apple and google map(if installed, Otherwise system will navigate to default apple map).
Add this OpenMaps class in your project.
class OpenMaps {
// Presenting Action Sheet with Map Options Like Google And Apple Maps.
/// - Parameter coordinate: coordinate of destination
static func presentActionSheetwithMapOption(coordinate: CLLocationCoordinate2D) {
let latitude = coordinate.latitude
let longitude = coordinate.longitude
// Google MAP URL
let googleURL = "comgooglemaps://?daddr=\(latitude),\(longitude)&directionsmode=driving"
let googleItem = ("Google Maps", URL(string:googleURL)!)
// Apple MAP URL
let appleURL = "maps://?daddr=\(latitude),\(longitude)"
var installedNavigationApps = [("Apple Maps", URL(string:appleURL)!)]
if UIApplication.shared.canOpenURL(googleItem.1) {
installedNavigationApps.append(googleItem)
}
if installedNavigationApps.count == 1 {
if let app = installedNavigationApps.first {
self.openMap(app: app)
}
return
}
// If there are google map install in the device then System will ask choose map application between apple and google map.
let alert = UIAlertController(title: nil,
message: "Choose application",
preferredStyle: .actionSheet)
for app in installedNavigationApps {
let button = UIAlertAction(title: app.0,
style: .default, handler: { _ in
self.openMap(app: app)
})
alert.addAction(button)
}
let cancel = UIAlertAction(title: "Cancel", style: .cancel, handler: nil)
alert.addAction(cancel)
guard let vc = AlertViewManager.shared.getTopViewController() else {
debugPrint("Unable To open the action sheet for maps. Function: \(#function), Line: \(#line)")
return
}
vc.present(alert, animated: true)
}
// Open Selected Map
/// - Parameter app: Selected Map Application Details
private static func openMap(app: (String, URL)) {
guard UIApplication.shared.canOpenURL(app.1) else {
debugPrint("Unable to open the map.")
return
}
UIApplication.shared.open(app.1, options: [:], completionHandler: nil)
}
}
Usage:
let coordinate = CLLocationCoordinate2D(latitude: 30.7333,
longitude: 76.7794)
OpenMaps.presentActionSheetwithMapOption(coordinate: coordinate)
Using Swift 5 and XCode 13, I was able to use Google's Universal cross-platform syntax to open Google Maps app in directions mode.
The link will open the google maps app if it is installed on the phone. If maps is not installed it will open the url in the browser.
let url = "https://www.google.com/maps/dir/?api=1&destination=\(coords.latitude)%2C\(coords.longitude)"
guard let googleUrl = URL.init(string: url) else {
// handle error
return
}
UIApplication.shared.open(googleUrl)
In Swift 4:
if (UIApplication.shared.canOpenURL(NSURL(string:"comgooglemaps://")! as URL)) {
UIApplication.shared.open((NSURL(string:
"comgooglemaps://?saddr=&daddr=\(trashDictionarySorted[indexPath.section][0])&directionsmode=driving")! as URL), options: [:], completionHandler: nil)
} else {
NSLog("Can't use comgooglemaps://");
}
}
The Answer is already there but in older versions of Swift, And this code can open Google Map in browser if your iPhone did't install Google Map Application.
In Swift 4
import MapKit
func openGoogleDirectionMap(_ destinationLat: String, _ destinationLng: String) {
let LocationManager = CLLocationManager()
if let myLat = LocationManager.location?.coordinate.latitude, let myLng = LocationManager.location?.coordinate.longitude {
if let tempURL = URL(string: "comgooglemaps://?saddr=&daddr=\(destinationLat),\(destinationLng)&directionsmode=driving") {
UIApplication.shared.open(tempURL, options: [:], completionHandler: { (isSuccess) in
if !isSuccess {
if UIApplication.shared.canOpenURL(URL(string: "https://www.google.co.th/maps/dir///")!) {
UIApplication.shared.open(URL(string: "https://www.google.co.th/maps/dir/\(myLat),\(myLng)/\(destinationLat),\(destinationLng)/")!, options: [:], completionHandler: nil)
} else {
print("Can't open URL.")
}
}
})
} else {
print("Can't open GoogleMap Application.")
}
} else {
print("Prease allow permission.")
}
}
func openTrackerInBrowser(lat: String, long: String, dlat: String, dlong: String){
if let urlDestination = URL.init(string: "https://www.google.co.in/maps/dir/?saddr=\(lat),\(long)&daddr=\(dlat),\(dlong)&directionsmode=driving") {
UIApplication.shared.open(urlDestination, options: [:], completionHandler: nil)
}
}

calling method in swift [duplicate]

I have converted the code for making a phone call from Objective-C to Swift, but in Objective-C, we can set the type of the URL that we like to open (e.g. telephone, SMS, web) like this:
#"tel:xx"
#"mailto:info#example.es"
#"http://stackoverflow.com"
#"sms:768number"
The code in Swift is:
UIApplication.sharedApplication().openURL(NSURL(string : "9809088798")
I read that have not released any scheme parameter for tel:, but I don't know if Swift can detect if the string is for making a phone call, sending email, or opening a website. Or may I write:
(string : "tel//:9809088798")
?
I am pretty sure you want:
UIApplication.sharedApplication().openURL(NSURL(string: "tel://9809088798")!)
(note that in your question text, you put tel//:, not tel://).
NSURL's string init expects a well-formed URL. It will not turn a bunch of numbers into a telephone number. You sometimes see phone-number detection in UIWebView, but that's being done at a higher level than NSURL.
EDIT: Added ! per comment below
A self-contained solution in Swift:
private func callNumber(phoneNumber:String) {
if let phoneCallURL:NSURL = NSURL(string:"tel://\(phoneNumber)") {
let application:UIApplication = UIApplication.sharedApplication()
if (application.canOpenURL(phoneCallURL)) {
application.openURL(phoneCallURL);
}
}
}
Now, you should be able to use callNumber("7178881234") to make a call.
For Swift in iOS:
var url:NSURL? = NSURL(string: "tel://9809088798")
UIApplication.sharedApplication().openURL(url!)
You need to remember to remove the whitespaces or it won't work:
if let telephoneURL = NSURL(string: "telprompt://\(phoneNumber.stringByReplacingOccurrencesOfString(" ", withString: ""))") {
UIApplication.sharedApplication().openURL(telelphoneURL)
}
"telprompt://" will prompt the user to call or cancel while "tel://" will call directly.
# confile:
The problem is that your solution does not return to the app after the
phone call has been finished on iOS7. – Jun 19 at 13:50
&# Zorayr
Hm, curious if there is a solution that does do that.. might be a
restriction on iOS.
use
UIApplication.sharedApplication().openURL(NSURL(string: "telprompt://9809088798")!)
You will get a prompt to Call/Cancel but it returns to your application. AFAIK there is no way to return (without prompting)
You must insert "+"\ is another way
private func callNumber(phoneNumber:String) {
if let phoneCallURL:NSURL = NSURL(string:"tel://"+"\(phoneNumber)") {
let application:UIApplication = UIApplication.sharedApplication()
if (application.canOpenURL(phoneCallURL)) {
application.openURL(phoneCallURL);
}
}
}
Small update to Swift 3
UIApplication.shared.openURL(NSURL(string: "telprompt://9809088798")! as URL)
The following code snippet can tell if the SIM is there or not and if the device is capable of making the call and if ok then it'll make the call
var info = CTTelephonyNetworkInfo()
var carrier = info.subscriberCellularProvider
if carrier != nil && carrier.mobileNetworkCode == nil || carrier.mobileNetworkCode.isEqual("") {
//SIM is not there in the phone
}
else if UIApplication.sharedApplication().canopenURL(NSURL(string: "tel://9809088798")!)
{
UIApplication.sharedApplication().openURL(NSURL(string: "tel://9809088798")!)
}
else
{
//Device does not have call making capability
}
For making a call in swift 5.1, just use the following code: (I have tested it in Xcode 11)
let phone = "1234567890"
if let callUrl = URL(string: "tel://\(phone)"), UIApplication.shared.canOpenURL(callUrl) {
UIApplication.shared.open(callUrl)
}
Edit:
For Xcode 12.4, swift 5.3, just use the following:
UIApplication.shared.open(NSURL(string: "tel://555-123-1234")! as URL)
Make sure that you import UIKit, or it will say that it cannot find UIApplication in scope.
For swift 3
if let phoneCallURL:URL = URL(string:"tel://\(phoneNumber ?? "")") {
let application:UIApplication = UIApplication.shared
if (application.canOpenURL(phoneCallURL)) {
application.open(phoneCallURL, options: [:], completionHandler: nil);
}
}
For swift 4:
func call(phoneNumber: String) {
if let url = URL(string: phoneNumber) {
if #available(iOS 10, *) {
UIApplication.shared.open(url, options: [:],
completionHandler: {
(success) in
print("Open \(phoneNumber): \(success)")
})
} else {
let success = UIApplication.shared.openURL(url)
print("Open \(phoneNumber): \(success)")
}
}
}
Then, use the function:
let phoneNumber = "tel://+132342424"
call(phoneNumber: phoneNumber)
Swift 4 and above
let dialer = URL(string: "tel://5028493750")
if let dialerURL = dialer {
UIApplication.shared.open(dialerURL)
}
I prefer deferring the URL creation to the built-ins like:
var components = URLComponents()
components.scheme = "tel"
components.path = "1234567890"
print(components.url!) //Prints out: "tel:1234567890"
Which can then be used in UIApplication.shared.openURL

How to share an Audio/Image to whatsapp from an iOS app using Swift

I would like to make an app that shares audio/photos to Whatsapp. How can I achieve this function. When i'm pressing share button audio/photo should download from remote server and needs to share to whatsapp contacts.
Im doing the app in swift.
Check this link: https://faq.whatsapp.com/en/iphone/23559013
This would be an example function in Swift 3.0:
func sendWhatsApp() {
let date = Date()
let msg = "Hi my dear friends\(date)"
let urlWhats = "whatsapp://send?text=\(msg)"
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed) {
if let whatsappURL = URL(string: urlString) {
if UIApplication.shared.canOpenURL(whatsappURL) {
UIApplication.shared.open(whatsappURL, options: [:], completionHandler: nil)
} else {
print("please install watsapp")
}
}
}
}

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