I want to open Google calendar app more specifically the create event page in the app from my another flutter app.
I am using the URL launcher package but it opens the app in chrome
What change should I make in the URL so that the add event page opens directly in the google calendar app.
Currently my URL looks like below
https://calendar.google.com/calendar/u/0/r/eventedit?dates=20210226T033000/20210226T040000&ctz=Asia/Calcutta&location&text=Blawsome:+A+Crystal+Alchemy+Healing+Meditation&details=Parth+Pitroda
My code for that part is as below
if (await canLaunchUrl(Uri.parse('https://calendar.google.com/calendar/u/0/r/eventedit?dates=20210226T033000/20210226T040000&ctz=Asia/Calcutta&location&text=Blawsome:+A+Crystal+Alchemy+Healing+Meditation&details=Parth+Pitroda'))) {
await launchUrl(
Uri.parse('https://calendar.google.com/calendar/u/0/r/eventedit?dates=20210226T033000/20210226T040000&ctz=Asia/Calcutta&location&text=Blawsome:+A+Crystal+Alchemy+Healing+Meditation&details=Parth+Pitroda'),
mode: LaunchMode.externalApplication);
} else {
throw 'Could not launch URL';
}
rathe than using a HTTP request, you could use googleapis package from https://pub.dev which has inbuilt methods to create Google Calender events directly within the Calender app.
Check package and documentation here.
Happy coding!
Related
I just released my first app and It has a button in it that takes you to a website.
A user just sent me this:.
I tried googling Google's secure browsers policy but not much info is coming up.
how can I make my app comply with this policy? I think the button opens a browser in app (I use duckduckgo as my default browser and haven't had an issue)
is it just a case of opening a browser and then heading to the website when the button is pressed?
my code to open the website is:
_launchURL() async {
const url = 'https://www.thiswebsite.com';
final uri = Uri.parse(url);
if (await canLaunchUrl(uri)) {
await launchUrl(uri);
} else {
throw 'Could not launch $url';
}
}
thanks so much and any help would be greatly appreciated
Google is trying to make sure, you open this window in an actual new browser window, not in a webview still under the control of your application.
Your code should open an external browser.
Maybe the user has no browser installed on their device? Maybe their default browser is some exotic thing not recognized by Google?
If you are using the latest version of url_launcher (currently 6.1.8) there is not a lot more you can do.
You could force the app to take the external browser, not the in-app webview:
await launchUrl(_url,mode: LaunchMode.externalApplication);
But that should be what happens anyway. If your version is up to date, ask your user, what browser they use. Be prepared to tell them that they need to use another one.
I have developed a Flutter Web app that uses Firebase Authentication in order to sign in users to the app.
I've declared the Firebase Authentication persistence field so that the app will remember and auto-login the user when he revisits the Flutter Web app's URL, and won't be required to re-login every time he launches the URL.
It all works fine on a regular browser, but when the user generates a PWA (for example, clicking "Add to Home Screen" on iOS devices to save the website as PWA), the persistence feature stops working, and the user is required to re-login every time he opens the PWA.
Is there a way to add Firebase Authentication's persistence feature to a PWA? And if not, is there a way to prevent generating a PWA (and saving the Flutter Web app as a regular browser URL when clicking "Add to Home Screen" button on iOS, for example)?
Thank you!
To solve the persistence problem, add a listener:
FirebaseAuth.instance.idTokenChanges().listen((User? user) async {
if (user == null) {
// Function for user not logged in here. Do not write function to change page here.
} else {
// As it's a Future it will take a while to process the user's information, so it
will call the function after it's done.
Navigator.pushReplacement(
context, MaterialPageRoute(builder: (_) => Home()));
}
}
This is an example I made and it worked, use controllers to change the status, put some function to wait for the information to be processed.
Hope this helps. Any questions, at your disposal.
I would like to create an app in Flutter. The web version contains a button that should open version of android or IOS app according user platform if mobile version of app was installed (like an app install or open banner).
How should I detect is app installed in web flutter?
update:
I tried below code using import 'package:universal_html/html.dart' pakage:
window.location.href = (defaultTargetPlatform ==
TargetPlatform.android)
? 'https://play.google.com/store/apps/details?id=com.amazon.mShop.android.shopping'
: 'https://apps.apple.com/us/app/amazon-shopping/id297606951';
But this just open the store. I'm looking for a solution to open app directly if it was installed.
If this is fine for you, you can use an URL launcher. This way it opens the App store or play store and the user can either download the App or open it.
For Example flutter has a package that does most of this work:
https://github.com/Purus/launch_review
LaunchReview.launch(androidAppId: "yourpackagename", iOSAppId: "appid");
You just need to pass your package name and on ios your app ID
You could also use an URL Launcher:
https://pub.dev/packages/url_launcher
The code would be similar to this:
_launchURL(String url) async {
if (await canLaunch(url)) {
await launch(url);
}
else {
throw 'Could not launch $url';
}
}
URL Example
try {
launch("market://details?id=" + appPackageName);
} on PlatformException catch(e) {
launch("https://play.google.com/store/apps/details?id=" + appPackageName);
} finally {
launch("https://play.google.com/store/apps/details?id=" + appPackageName);
}
Note this code needs to be adapted
Also see this tutorial for help: https://flutteragency.com/open-appstore-playstore-url-in-flutter/
Edit:
If you want to directly open another app you can use something like this:
https://pub.dev/packages/external_app_launcher/
flutter pub add external_app_launcher
The Code would look like this then:
await LaunchApp.openApp(
androidPackageName: 'net.pulsesecure.pulsesecure',
iosUrlScheme: 'pulsesecure://',
appStoreLink: 'itms-apps://itunes.apple.com/us/app/pulse secure/id945832041',// openStore: false
);
// Enter the package name of the App you want to open and for iOS add the URLscheme to the Info.plist file.
// The `openStore` argument decides whether the app redirects to PlayStore or AppStore.
// For testing purpose you can enter com.instagram.android
More infos regarding implementation and additional setup infos you can find here: https://pub.dev/packages/external_app_launcher in the Readme
What should i define in my activity i.e Action, ActivityClass, ActivityPackage, ExtraKey and ExtraValue to define an intent in app to open Youtube app in and android phone to open with "ExtraValue" searched in youtube?
I am trying the following combination but getting Error 601.
Action: android.intent.action.SEARCH
ActivityClass: com.google.android.youtube.WatchActivity
ActivityPackage: com.google.android.youtube
DataType:
DataUri:
ExtraKey: query
ExtraValue: app Inventor activity starter
ResultName:
use this on your project:
https://developers.google.com/youtube/android/player/downloads/
include this code in your onclick method:
Intent intent = YouTubeStandalonePlayer.createVideoIntent(getActivity(), api_key, video_id);
startActivity(intent);
api_key is the Google Console APIs developer key.
video_id is the code after v=, for example:fYWZNN4bbz8 in "https://www.youtube.com/watch?v=fYWZNN4bbz8"
:)
I need a web and mobile application which will work on android, iphone and windows as well.
I want to use facebook login for my apps and customer can like my page after login thats why i am using cordova to convert my web app into android app but i need to integrate facebook-like option which i didn't get in plugin.
So i implemented this using oglike
$scope.facebookLike = function() {
if($localStorage.hasOwnProperty("accessToken") === true) {
$http.post("https://graph.facebook.com/me/og.likes?access_token="+$localStorage.accessToken+"&object=https://www.facebook.com/Nxtlife-Technologies-Ltd-UK-180614345644169/?ref=br_rs");
alert("like sucessfully done");
} else {
alert("Not signed in");
$scope.facebookLogin();
}
}
Problem:
The code will not throwing any error but after success message it didin't make any changes to my page. like count is remains same.
og.likes is for Open Graph objects – external URLs, outside of Faceook.
You can not like Facebook pages by any other means than the official Like button.