In flutter code to open maps is like this. It is working on google but not launching apple map url. Might be small mistake or missing something
String addrs = '';
addrs = '###';
String googleUrl =
'comgooglemaps://?q=${Uri.encodeComponent(addrs)}¢er=lat,lng';
String appleUrl =
'https://maps.apple.com/?q=${addrs}&sll=lat,lng';
if (await canLaunch("comgooglemaps://")) {
print('launching com googleUrl');
await launch(googleUrl);
} else if (await canLaunch(appleUrl)) {
print('launching apple url');
await launch(appleUrl);
} else {
print('Could not launch url');
}```
Please encode the URL for apple maps as you have done for google maps.
Related
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.
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!
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 $url';
}
}
var authenticator = Authenticator(client,
scopes: scopes, port: 4000, urlLancher: urlLauncher);
var c = await authenticator.authorize();
closeWebView();
print(await c.getUserInfo());
}catch (err) {
print(err);
}
}
But I have no idea what to give as the uri, clientId and scopes.
For the uri I've tried https://steamcommunity.com/openid but when I try this I'm getting an error that it tried to Json.parse a xml response.
I've also tried the example for the steam_login package here, but that does nothing and doesn't feel like it should work within a Flutter app (even though the package states it Flutter compatible).
So if anybody could explain to me how I could achieve this with a working sample that would be great.
If I get this to work I'm planning on creating a tutorial/guide for it for others.
Looking for a good sample working code for downloading any file from URL in flutter. If its with native downloader then this will be very good. Please help me with sample of code to download any file using native downloader in flutter.
I have used few libraries but didn't turned out well.
For a mobile device, I used the http package to download a file to the applications document directory
First, create a httpClient object
static var httpClient = new HttpClient();
Then you can create a function like this to download the file:
Future<void> _downloadFile({
required String fileName,
}) async {
String url = ...;
var request = await httpClient.getUrl(Uri.parse(url));
var response = await request.close();
var bytes = await consolidateHttpClientResponseBytes(response);
String dir = (await getApplicationDocumentsDirectory())!.path;
File file = new File('$dir/$fileName'); // Note: Filename must contain the extension of the file too, like pdf, jpg etc.
await file.writeAsBytes(bytes);
}
For a flutter web application, I felt the url_launcher package was the easiest to work with.
_launchURL() async {
String url = ...;
if (await canLaunch(url)) {
await launch(url);
print('URL Launcher success');
} else {
throw Exception('Could not launch $url');
}
}
The url_launcher package code works even for a mobile device but it opens a new browser window to download the required file which is not a good user experience so I have used 2 approaches for the same problem.
Hope I have answered your query.
I want to convert path "content://media/external/images/media/138501" to File and set in the Image.
Code:
File imageFile = File("content://media/external/images/media/138501");
is not working on:
DecorationImage(image: ExactAssetImage(imageFile.path),fit: BoxFit.fill)
You can use uri_to_file package. It supports content:// URI.
Simple to use (Updated)
import 'package:uri_to_file/uri_to_file.dart';
try {
String uriString = 'content://sample.txt';
// Don't pass uri parameter using [Uri] object via uri.toString().
// Because uri.toString() changes the string to lowercase which causes this package to misbehave
// If you are using uni_links package for deep linking purpose.
// Pass the uri string using getInitialLink() or linkStream
File file = await toFile(uriString);
} on UnsupportedError catch (e) {
print(e.message);
} on IOException catch (e) {
print(e);
} catch (e) {
print(e);
}
then you can use this file as you want
Important note
Don't pass uri value like this
File file = await toFile(uri.toString());
Use like this if you are using uni_links package for deep linking purpose. Use getInitialLink() or linkStream
String? uriString = await getInitialLink();
if (uriString != null) {
File file = await toFile(uriString);
}
linkStream.listen((uriString) async {
if (uriString != null) {
File file = await toFile(uriString!);
}
});
This will not modify the uri string as we are not using uri.toString()
So this package will work fine
Working example:
Working example
For more details: uri_to_file
According to the doc you can use the fromUri constructor: File.fromUri(Uri uri).
Your case:
File imageFile = File.fromUri("content://media/external/images/media/138501");
You can use Absolute path plugin.
final filePath = await FlutterAbsolutePath.getAbsolutePath(uriString);
and then show the image by using Image.file:
Image.file(
File(filePath),
);
Make use of flutter_absolute_path package.
flutter_absolute_path: ^1.0.6 In pubsec.yaml
To convert file path from this format : “content://media/external/images/media/5275” To
"/storage/emulated/0/DCIM/Camera/IMG_00124.jpg”
final filePath = await FlutterAbsolutePath.getAbsolutePath("Any path
here");
File tempFile = File(filePath);
if (tempFile.existsSync()) {
//add your logic here.
}
define your URL:
final url = "Your Image Url";
then convert the Image Url to File in flutter :
XFile _xFile = XFile(url);
Done.
You could use Image.file constructor.
DecorationImage(
image: Image.file(File("content://media/external/images/media/138501")),
fit: BoxFit.fill
)
Please take a note: On Android, this may require the android.permission.READ_EXTERNAL_STORAGE permission.