Yes you can; it works on Android too: tel: phone_number Calls the entered phone number. Valid telephone numbers as defined in the IETF RFC 3966 are accepted. Valid examples include the following: * tel:2125551212 * tel: (212) 555 1212 The Android browser uses the Phone app to handle the “tel” scheme, as defined by RFC … Read more
Update with Android 8.0 Oreo Even though the question was originally asked for Android L support, people still seem to be hitting this question and answer, so it is worth describing the improvements introduced in Android 8.0 Oreo. The backward compatible methods are still described below. What changed? Starting with Android 8.0 Oreo, the PHONE … Read more
Call the launch method from url_launcher package: launch(“tel://214324234”); Here’s the complete code: import ‘package:flutter/material.dart’; import ‘package:url_launcher/url_launcher.dart’; class MyApp extends StatelessWidget { @override Widget build(BuildContext context) { return new MaterialApp( title: ‘Flutter Demo’, home: new Home(), ); } } class Home extends StatelessWidget { Home({Key key}) : super(key: key); @override Widget build(BuildContext context) => new Scaffold( … Read more
You can call like this: if let url = URL(string: “tel://\(number)”) { UIApplication.shared.openURL(url) } For Swift 3+, you can use like guard let number = URL(string: “tel://” + number) else { return } UIApplication.shared.open(number) OR UIApplication.shared.open(number, options: [:], completionHandler: nil) Make sure you’ve scrubbed your phone number string to remove any instances of (, ), … Read more
Just try: if let url = NSURL(string: “tel://\(busPhone)”) where UIApplication.sharedApplication().canOpenURL(url) { UIApplication.sharedApplication().openURL(url) } assuming that the phone number is in busPhone. NSURL‘s init(string:) returns an Optional, so by using if let we make sure that url is a NSURL (and not a NSURL? as returned by the init). For Swift 3: if let url = … Read more
If you look at the source code for react-native-phone-call, it’s ultimately just a wrapper for: import {Linking} from ‘react-native’ Linking.openURL(`tel:${phoneNumber}`)
use a PhoneStateListener to see when the call is ended. you will most likely need to trigger the listener actions to wait for a the call to start (wait until changed from PHONE_STATE_OFFHOOK to PHONE_STATE_IDLE again) and then write some code to bring your app back up on the IDLE state. you may need to … Read more
You forgot to call startActivity. It should look like this: Intent intent = new Intent(Intent.ACTION_CALL); intent.setData(Uri.parse(“tel:” + bundle.getString(“mobilePhone”))); context.startActivity(intent); An intent by itself is simply an object that describes something. It doesn’t do anything. Don’t forget to add the relevant permission to your manifest: <uses-permission android:name=”android.permission.CALL_PHONE” />
To go back to original app you can use telprompt:// instead of tel:// – The tell prompt will prompt the user first, but when the call is finished it will go back to your app: NSString *phoneNumber = [@”telprompt://” stringByAppendingString:mymobileNO.titleLabel.text]; [[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
You can use Intent.ACTION_DIAL instead of Intent.ACTION_CALL. This shows the dialer with the number already entered, but allows the user to decide whether to actually make the call or not. ACTION_DIAL does not require the CALL_PHONE permission.