I want to make the container click lead to a specific youtube url which should open in the the youtube app and not in webview inside the app.
add in pubspec.yaml:
dependecies:
url_launcher: ^5.4.2
add this in your code:
LaunchUrl('https://www.youtube.com');//or any link you want
You should look into the package android_intent. Basically it will resolve the url if it has been defined as a valid link for an application in your android parameters (which is enabled by default for youtube).
Code Sample
AndroidIntent intent = AndroidIntent(
action: 'action_view',
data: url,
);
await intent.launch();
you can use package url_launcher with the function below:
launchUrl(
Uri.parse("http://www.youtube.com/watch?v=${youtubeID}"),
mode: LaunchMode.externalApplication);
Related
I want to open Google calendar app more specifically the create event page in the app from my another flutter app.
I am using the URL launcher package but it opens the app in chrome
What change should I make in the URL so that the add event page opens directly in the google calendar app.
Currently my URL looks like below
https://calendar.google.com/calendar/u/0/r/eventedit?dates=20210226T033000/20210226T040000&ctz=Asia/Calcutta&location&text=Blawsome:+A+Crystal+Alchemy+Healing+Meditation&details=Parth+Pitroda
My code for that part is as below
if (await canLaunchUrl(Uri.parse('https://calendar.google.com/calendar/u/0/r/eventedit?dates=20210226T033000/20210226T040000&ctz=Asia/Calcutta&location&text=Blawsome:+A+Crystal+Alchemy+Healing+Meditation&details=Parth+Pitroda'))) {
await launchUrl(
Uri.parse('https://calendar.google.com/calendar/u/0/r/eventedit?dates=20210226T033000/20210226T040000&ctz=Asia/Calcutta&location&text=Blawsome:+A+Crystal+Alchemy+Healing+Meditation&details=Parth+Pitroda'),
mode: LaunchMode.externalApplication);
} else {
throw 'Could not launch URL';
}
rathe than using a HTTP request, you could use googleapis package from https://pub.dev which has inbuilt methods to create Google Calender events directly within the Calender app.
Check package and documentation here.
Happy coding!
await launchUrl(
Uri.parse(
'https://www.facebook.com/$fbid',
),
mode: LaunchMode.externalApplication);
I have this launch url mode. The address works when I use it on my browser however, it does not when I use it inside the Flutter app. It says The link you followed may be broken, or ....
Twitter works fine but Instagram and FB are the ones that cause the error. I feel like something must be done for meta apps that I am missing right now. How can I fix this?
if (Platform.isAndroid) { var fbUrl = "fb://facewebmodal/f?href=" + "https://www.facebook.com/$fbid "; //for android launchFacebook(fbUrl, "https://www.facebook.com/$fbid"); }
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
I am using flutter_webview_plugin: ^0.3.11
This is my code
Widget build(BuildContext context){
return WebviewScaffold(
url: glbPhotoURL,
withJavascript: true,
scrollBar : true,
withZoom: true
)
url: glbPhotoURL => here glbPhotoURL is a URL that I am passing
When I am using any normal URL it is running fine (like http://www.google.com, http://youtube.com"
Even url like - https://youtu.be/o5UPfG1eIw4 is running fine
But when I am using any google photo url (short url) it is throwing an error net::ERR_UNKNOWN_URL_SCHEME for eg - https://photos.app.goo.gl/FkQenAD8kQQc4TSr6
If I am using the expanded URL it shows the pictures -https://photos.google.com/share/AF1QipNItZG3Cg_hn9__2QnuVh3nNMbRuGxQaQSWZ76qni7L7h0ORbauolcH3AKe0MOnEA?
key=emc1Mk1CenRJRjloMjV5V1AzcmczNUprcGFsbmR3
Please help me resolve the issue
As of now I am running it on Android physical device.
Google Photos uses Firebase Dynamic Links. I suggest launching the link externally. I encountered a similar error on Android before, when Firebase Dynamic Links are being forced to be loaded in a WebView. FDLs are expected to be handled by Google Play Services in Android. But since the WebView doesn't know what to do with the link it's forced to display, the WebView returns "net::ERR_UNKNOWN_URL_SCHEME" error.
Open the link externally by using url_launcher. Use RegEx to filter intent URLs and check if the URL can be launched and be handled externally (outside the app).
var yourURL = "URL goes here";
// Check if URL contains Google Photos URL
yourURL.contains(RegExp('^https://photos\.app\.goo\.gl/.*$')){
// Check if the URL can be launched
if (await canLaunch(yourURL)) {
await launch(yourURL);
} else {
print('Could not launch $yourURL');
}
}
url_launcher helps to draft an email, make a call and send a sms to a specific recipient. Is there any way to launch mail/sms apps (inbox - not draft) ?
I'm not sure if you can do that in iOS but for Android, use android_intent plugin and put the appropriate package_name in data.
if (platform.isAndroid) {
AndroidIntent intent = AndroidIntent(
action: 'action_view',
data: 'package:com.example.app',
);
await intent.launch();
}