Flutter Dynamic linking not available for web apps? - flutter

My Use Case:
Users in my app can create lists of items. I want the users to be able to share a link to one of their lists.
I want individuals who click the shared link to see the list of items in a web browser and not need the app installed.
Create the Dynamic Link
The first thing I need to do is create a dynamic link for a user to share. I struggled with this at first and then looked at this question here: Flutter - How to pass custom arguments in firebase dynamic links for app invite feature?
After looking at that I build my own class like so:
class PackDynamicLinkService {
Future<Uri> createDynamicLink({required String packID}) async {
final DynamicLinkParameters parameters = DynamicLinkParameters(
// This should match firebase but without the username query param
uriPrefix: 'ttps://appName.app/p',
// This can be whatever you want for the uri, https://yourapp.com/groupinvite?username=$userName
link: Uri.parse('https://appName.app/p?pack=$packID'),
androidParameters: const AndroidParameters(
packageName: 'com.test.demo',
minimumVersion: 1,
),
iosParameters: const IOSParameters(
bundleId: 'com.test.demo',
minimumVersion: '1',
appStoreId: '',
),
);
final dynamicLink = await FirebaseDynamicLinks.instance.buildShortLink(
parameters,
shortLinkType: ShortDynamicLinkType.unguessable,
);
return dynamicLink.shortUrl;
}
}
I then called that function in my app and I am able to get a URL that looks like this: https://appName.app/p?pack=lVeSBUHoLX52zPqAiB9A
So far so good. Now I just need to click the link and read the parameter $packID that I passed to it inside my web app. Then I can route them to the correct view and display the list.
Processing Dynamic Link
in my marin.dart I tried to retrieve the params with:
//get link stuff
final PendingDynamicLinkData? initialLink =
await FirebaseDynamicLinks.instance.getInitialLink();
if (initialLink != null) {
final Uri fullLink = initialLink.link;
print(fullLink.queryParameters['pack']);
}
but when I upload to firebase hosting to see if this code works for testing, I get this error:
Uncaught MissingPluginException(No implementation found for method
FirebaseDynamicLinks#getInitialLink on channel
plugins.flutter.io/firebase_dynamic_links)
After doing some more research it seems that firebase_dynamic_links package may not be available for web? So does that mean my original use case is not possible with flutter at this time?

You would want to take the link received and parse it using something like:
var PendingDynamicLinkData? initialLink =
await FirebaseDynamicLinks.instance.getInitialLink();
if (kIsWeb) {
String link = Uri.base;
initialLink = await
FirebaseDynamicLinks.instance.getDynamicLink(Uri.parse(link));
}
if (initialLink != null) {
final Uri fullLink = initialLink.link;
print(fullLink.queryParameters['pack']);
}
Here, we are checking if its a web app and then we can parse the dynamic link and get the info for it. We just check for the Uri that was used in our web app.

Related

Custom Parameters are always Null | Firebase Dynamic Links

What I Want to Do
When the user taps the URL, it launches the app and pops up a page within the app based on the included custom parameters.
Problem
I succeeded in launching an Android or iOS app by tapping the URL, but the custom parameter is always null.
What I Did
The URL that users tap is as follows.
https://example.page.link/test/?link=https://test.example.com/?id=${id}&apn=com.example.app
id is custom parameter that I want to get when this URL is called.
I created link on Firebase Console like example.page.link/test.
In initState, call initDeepLinks function.
// openPage opens detail page with id
Future<void> initDeepLinks() async {
final PendingDynamicLinkData? initialLink =
await FirebaseDynamicLinks.instance.getInitialLink();
if (initialLink != null) {
final Uri deepLink = initialLink.link;
print("A: " + deepLink.queryParameters["id"]!);
openPage(deepLink.queryParameters["id"]!);
}
FirebaseDynamicLinks.instance.onLink.listen((dynamicLinkData) {
print("B: " + dynamicLinkData.link.queryParameters["id"]!);
openPage(dynamicLinkData.link.queryParameters["id"]!);
}).onError((error) {});
}
But those two print()s print only A: or B: .
I checked URL that contains ?id=$id and it is correct.
Tried other URL
When I change the URL to
https://example.page.link/?link=https://test.example.com/?id=${id}&apn=com.example.app
(removed `/test` path)
It works with custom parameters only on Android.
But do same on iOS, the browser shows https://test.example.com page and app doesn't appear.
What should I do to make custom parameters?

How to pass parameters to a dynamic link from flutter?

I want to navigate a user to a particular page on clicking a shared link based on the id passed in that link. I use dynamic link of Firebase for that. after configuring a dynamic link in the Firebase console manually. I use this code to create a link when sharing an app:
Future<Uri> createDynamicLink(String id) async {
final DynamicLinkParameters parameters = DynamicLinkParameters(
uriPrefix: 'https://asbdf.page.link',
link: Uri.parse('https://aspdf.page.link/createpost?id=$id'),
androidParameters: AndroidParameters(
packageName: 'com.sample.aspdf',
minimumVersion: 1,
),
);
print(parameters.link);
final ShortDynamicLink shortLink = await dynamicLinks.buildShortLink(parameters);
final Uri shortUrl = shortLink.shortUrl;
return shortUrl;
}
The sharing link works fine. After sharing the link on social media (whatsapp), when I click on that link, the app opens and causes an error. This is the code to retrieve deeplink:
final Uri deepLink = data.link;
print("deeplink found");
print(deepLink);
String? id = deepLink.queryParameters['id'];
print(id.toString());
Get.offAll(()=>AuthStatus(uid: id!));
This is because the app is opening the static deeplink which I configured in the console manually, not the link which was created programatically from the above code. So no parameters are taken by the link and the value of the id variable becomes null and causes an error. How can I fix this problem?

Calendar clientViaUserConsent it gives me Authorization Error while creating event

clientViaUserConsent opens URL in browser but it said invalid request. this URL is generated internally from lib. I had double-checked my ClientId for both platforms but still face issues for getting AuthClient for create a calendar event.
I used the below packages to create events in the google calender.
googleapis: ^8.1.0
googleapis_auth: ^1.3.0
static final androidClientId = ClientId('xxxxxxxx.apps.googleusercontent.com');
static final iOSClientId = ClientId('xxxxxxx.apps.googleusercontent.com');
final _clientID = Platform.isAndroid ? EventProvider.androidClientId : EventProvider.iOSClientId;
final _scopes = [CalendarApi.calendarScope];
clientViaUserConsent(_clientID, _scopes, prompt).then((AuthClient client) {
var calendar = CalendarApi(client);
}
void prompt(String url) async {
print(" => $url");
if (await canLaunch(url)) {
await launch(URL);
} else {
throw 'Could not launch $url';
}
}
I am referring to this article for creating an event in google calendar.
https://blog.codemagic.io/google-meet-events-in-flutter/
https://medium.com/flutter-community/flutter-use-google-calendar-api-adding-the-events-to-calendar-3d8fcb008493
You are seeing that error because the app hasn't been verified. If you are the app developer I advise you to check the App Verification FAQ to learn more about the verification steps. If you aren't the developer, you could try to enable the less secure app access but please be mindful of the consequences:
Less secure apps can make it easier for hackers to get in to your account, so blocking sign-ins from these apps helps keep your account safe.

Flutter open whatsapp with text message

I want to open whatsapp from my Flutter application and send a specific text string. I'll select who I send it to when I'm in whatsapp.
After making some research I came up with this:
_launchWhatsapp() async {
const url = "https://wa.me/?text=Hey buddy, try this super cool new app!";
if (await canLaunch(url)) {
await launch(url);
} else {
throw 'Could not launch $url';
}
}
Which works ok ish, however there are two problems:
As soon as I make the text string into multi words it fails. So if I change it to:
_launchWhatsapp() async {
const url = "https://wa.me/?text=Hey buddy, try this super cool new app!";
if (await canLaunch(url)) {
await launch(url);
} else {
throw 'Could not launch $url';
}
}
Then the Could not launch $url is thrown.
I have whatsapp already installed on my phone, but it doesn't go directly to the app, instead it gives me a webpage first and the option to open the app.
Here is the webpage that I see:
Any help on resolving either of these issues would be greatly appreciated.
Thanks
Carson
P.s. I'm using the Url_launcher package to do this.
From the official Whatsapp FAQ, you can see that using "Universal links are the preferred method of linking to a WhatsApp account".
So in your code, the url string should be:
const url = "https://wa.me/?text=YourTextHere";
If the user has Whatsapp installed in his phone, this link will open it directly. That should solve the problem of opening a webpage first.
For the problem of not being able to send multi-word messages, that's because you need to encode your message as a URL. Thats stated in the documentation aswell:
URL-encodedtext is the URL-encoded pre-filled message.
So, in order to url-encode your message in Dart, you can do it as follows:
const url = "https://wa.me/?text=Your Message here";
var encoded = Uri.encodeFull(url);
As seen in the Dart Language tour.
Please note that in your example-code you have put an extra set of single quotes around the text-message, which you shouldn't.
Edit:
Another option also presented in the Whatsapp FAQ is to directly use the Whatsapp Scheme. If you want to try that, you can use the following url:
const url = "whatsapp://send?text=Hello World!"
Please also note that if you are testing in iOS9 or greater, the Apple Documentation states:
Important
If your app is linked on or after iOS 9.0, you must declare the URL schemes you pass to this method by adding the LSApplicationQueriesSchemes key to your app's Info.plist file. This method always returns false for undeclared schemes, whether or not an appropriate app is installed. To learn more about the key, see LSApplicationQueriesSchemes.
So you need to add the following keys to your info.plist, in case you are using the custom whatsapp scheme:
<key>LSApplicationQueriesSchemes</key>
<array>
<string>whatsapp</string>
</array>
till date: 27-06-2022
using package: https://pub.dev/packages/url_launcher
dependencies - url_launcher: ^6.1.2
TextButton(
onPressed: () {
_launchWhatsapp();
},
)
_launchWhatsapp() async {
var whatsapp = "+91XXXXXXXXXX";
var whatsappAndroid =Uri.parse("whatsapp://send?phone=$whatsapp&text=hello");
if (await canLaunchUrl(whatsappAndroid)) {
await launchUrl(whatsappAndroid);
} else {
ScaffoldMessenger.of(context).showSnackBar(
const SnackBar(
content: Text("WhatsApp is not installed on the device"),
),
);
}
}
Here is new update way...
whatsapp() async{
var contact = "+880123232333";
var androidUrl = "whatsapp://send?phone=$contact&text=Hi, I need some help";
var iosUrl = "https://wa.me/$contact?text=${Uri.parse('Hi, I need some help')}";
try{
if(Platform.isIOS){
await launchUrl(Uri.parse(iosUrl));
}
else{
await launchUrl(Uri.parse(androidUrl));
}
} on Exception{
EasyLoading.showError('WhatsApp is not installed.');
}
}
and call whatsapp function in onpress or ontap function
For using the wa.me domain, make sure to use this format...
https://wa.me/123?text=Your Message here
This will send to the phone number 123. Otherwise, you will get an error message (see? https://wa.me/?text=YourMessageHere ). Or, if you don't want to include the phone number, try this...
https://api.whatsapp.com/send?text=Hello there!
Remember, wa.me requires a phone number, whereas api.whatsapp.com does not. Hope this helps!
I know it is too late for answering this, but for those who want the same functionality, the current way is to do it like this:
launchUrl(Uri.parse('https://wa.me/$countryCode$contactNo?text=Hi'),
mode: LaunchMode.externalApplication);
if you will use URL Launcher then the whatsapp link will be open on web browser. So you need to set parameter - not to open on safari browser. The complete code you can find on this flutter tutorial.
But for your case use below code.
await launch(whatappURL, forceSafariVC: false);
Today i am adding solution its working fine on my desktop and phone
Add 91 if your country code is +91
Remember not add any http or https prefix otherwise wont work.
whatsapp://send?phone=9112345678&text=Hello%20World!

How to use Steam (OpenId) to login into my Firebase powered Flutter app?

I'm building a Flutter app and I want users to be able to login through Steam (OpenId).
The app is powered by Firebase.
I'm pretty new to the whole OpenId world and I've read into it but I'm still not getting it.
I found that Firebase Supports signinWithCustomToken so I think I should get the Steam OpenId to get me one of those.
I also found two packages which might help me but I couldn't get those to work either:
https://pub.dev/packages/openid_client
https://pub.dev/packages/steam_login
If tried the following sample from the openid_client package:
_authenticate(Uri uri, String clientId, List<String> scopes) async {
try {
var issuer = await Issuer.discover(uri);
var client = Client(issuer, clientId);
print('CLAIMS:');
print(issuer.claimsMap);
urlLauncher(String url) async {
if (await canLaunch(url)) {
await launch(url, forceWebView: true);
} else {
throw 'Could not launch $url';
}
}
var authenticator = Authenticator(client,
scopes: scopes, port: 4000, urlLancher: urlLauncher);
var c = await authenticator.authorize();
closeWebView();
print(await c.getUserInfo());
}catch (err) {
print(err);
}
}
But I have no idea what to give as the uri, clientId and scopes.
For the uri I've tried https://steamcommunity.com/openid but when I try this I'm getting an error that it tried to Json.parse a xml response.
I've also tried the example for the steam_login package here, but that does nothing and doesn't feel like it should work within a Flutter app (even though the package states it Flutter compatible).
So if anybody could explain to me how I could achieve this with a working sample that would be great.
If I get this to work I'm planning on creating a tutorial/guide for it for others.