I need to detect, if an app is running on an iOS or Android emulator to skip an QR code scan method and just return a scanned code.
Q: How do I detect
on which device type - iOS or Android - an app is running and
if an app is running on an emulator?
Just found this plugin, which prints various details:
https://pub.dartlang.org/packages/device_info#-readme-tab-
Output on Android emulator [see last line]:
safe_device: ^1.1.1
import 'package:safe_device/safe_device.dart';
Checks whether device is real or emulator
bool isRealDevice = await SafeDevice.isRealDevice;
(ANDROID ONLY) Check if development Options is enable on device
bool isDevelopmentModeEnable = await SafeDevice.isDevelopmentModeEnable;
I have problem sharing a link on Facebook in my app made with Appcelerator
I’m using iOS 11 Simulator, Titanium SDK 7.1 and Facebook Module 5.8.
My tiapp.xml should be correct.
The facebook app is not installed on the simulator.
My code is:
var fb = require('facebook');
fb.initialize();
function shareLink() {
fb.addEventListener('shareCompleted', onShareCompleted);
fb.presentShareDialog({
link: "http://www.google.com"
});
}
function onShareCompleted(e){
fb.removeEventListener('shareCompleted', onShareCompleted);
if (e.success) Ti.API.info('Share request succeeded.');
else Ti.API.info('Failed to share.' + JSON.stringify(e));
}
When I try to share, the app show me a webpage of facebook with the error “The parameter ‘href’ or ‘media’ is required”.
If I switch the Facebook module from version 5.8 to 5.6, the first time I try to share, it fails, but the second time it work!
I can’t understand where I’m wrong.
Thanks for any help!
Try this. Its super easy and I always use it for social sharing:
https://github.com/ricardoalcocer/socialshare/tree/master/app/widgets/com.alcoapps.socialshare
I am developing an app in ionic 3 for android and ios platform. For deeplinking, I am using cordova firebase dynamic link plugin.
This is working fine for android platform. But in the ios platform, though it is not throwing any error, it is not working.
My current ionic code.
this.platform.ready()
.then(() => {
return this.firebaseDynamicLinks.onDynamicLink();
}).then((dynamicLink:any) => {
console.log(dynamicLink); // always gives {matchType: "weak", deepLink:""} in ios
}).catch((error: any) => {
console.log(error);
});
The android part working fine. But in ios, the deeplink is always coming as empty.
my manually created dynamic url
https://xxxx.app.goo.gl/?link=<encoded url to mysite>&apn=<android bundle id>&ibi=<ios bundle id>
The provided link is returning HTTP 200 status when opening in the browser.
Testing steps (in ios):
In my iphone (ios9) I have placed this dynamic link in notepad
Then by clicking on the dynamic link the app opens.
In the console, I get
{matchType: "weak", deepLink:""}
I'm going through the docs in React Native and can only find navigating to external links from the app I am in.
I want to be able to navigate to the Settings app (more specifically to the privacy > location services page) but, can not seem to find the necessary information on it. There is the native iOS way of doing it which I am trying to replicate through React Native.
Is this possible?
I have tried the following to detect if there is a Settings URL. The console logs that the Settings url works however, it does not navigate to that page.
Update: thanks to #zvona I am now navigating to the settings page but not sure how to get to a specific deep link.
Linking.canOpenURL('app-settings:').then(supported => {
console.log(`Settings url works`)
Linking.openURL('app-settings:'
}).catch(error => {
console.log(`An error has occured: ${error}`)
})
You can access settings of the application with:
Linking.openURL('app-settings:');
But I don't know (yet) how to open a specific deep-link.
2021 update use:
Linking.openSettings();
otherwise your app will be rejected due use of non-public URL scheme
I successfully opened the settings by the code below, hope it's helpful :)
Linking.canOpenURL('app-settings:').then(supported => {
if (!supported) {
console.log('Can\'t handle settings url');
} else {
return Linking.openURL('app-settings:');
}
}).catch(err => console.error('An error occurred', err));
Reference: https://facebook.github.io/react-native//docs/linking.html
Since React Native version 0.59 this should be possible using openSettings();. This is described in the React Native Linking documentation. Although it did not work for me. When I tried quickly I saw a _reactNative.Linking.openSettings is not a function error message.
Linking.openSettings();
You can deep-link referencing the settings's index like so:
Linking.openURL('app-settings:{index}')
For example Linking.openURL('app-settings:{3}') would open the Bluetooth settings.
Linking.openURL('app-settings:1');
Adding an answer that worked for me and is easy to apply.
openSettings function in #react-native-community/react-native-permissions works for both iOS and Android.
Calling openSettings function will direct the user to the settings page of your app.
import { openSettings } from 'react-native-permissions';
openSettings();
You can import Linking from 'react-native' and then use Linking.openSettings() to trigger the call. This link explains it very concisely:
https://til.hashrocket.com/posts/eyegh79kqs-how-to-open-the-settings-app-in-reactnative-060
For example: to navigate under Settings/Bluetooth you have to use Linking.openURL('App-Prefs:Bluetooth');
For iOS 14 and ReactNative 16.13
You can use the most easiest way to open the app setting from react-native.
just,
import { Linking } from 'react-native';
and user below line anywhere you want open the app setting.
Linking.openSettings();
Old question, but this didn't work for me on Android and I found something that did. Hope this helps anyone looking for the same. :)
https://github.com/lunarmayor/react-native-open-settings
I don't have the ability to test it for iOS though.
Opens the platform specific settings for the given application.
You can handle your case using Linking from react-native. In my case, I accessed the touch id settings on IOS using:-
Linking.openURL('App-Prefs:PASSCODE');
Could not find information. I built pg app with 'cordova-plugin-wkwebview-engine'.
How to check if phonegap web app on ios 9.2 is really using WKWebview? Could not find info how userAgent string should look like for WKWebView.
Previous answer doesn't work on iOS 10, but you can use this code
if (window.webkit) {
//WKWebView
}
Old answer:
You can check for indexedDB
if (window.indexedDB) {
//WKWebView
}