I want to be able to open Skype with a call setup with a specific username from my react-native app. I had planned to use linking to do this, For example:
onLaunchPress = () => {
const { user } = this.props
Linking.openURL('https://skype.com/' + user)
.catch((error) => console.log(error))
})
This is not enough to add the correct query parameters, However I am unable to view the documentation on this as the page doesn't load here:
https://dev.skype.com/skype-uri
Any advice would be greatly appreciated, I am also concerned as to whether this is actually possible given what is posted here:
https://answers.microsoft.com/en-us/skype/forum/skype_web-skype_messms-skype_instamessms/open-skype-for-web-from-a-react-native-app/eef6a4c6-6b05-4566-b9e1-2f13f9d05606
According to this, you should be able to open Skype with the following:
Linking.openUrl('skype:');
You should also be able to pass the User in too with:
Linking.openUrl(`skype:${user}`);
Related
According to expos documentation, if I want user to navigate to another screen after user taps the notification received I should use deep linkedin. thats also the suggested method in all similar posts Ive seen. In my case I just do the following:
useEffect(()=> {
responseListener.current = Notifications.addNotificationResponseReceivedListener(response => {
if (response) {
navigation.navigate("Page 1")
}
});
}, [])
It works fine, when I have app open or closed. I wonder if I just dont understand something and there will be issues as everywhere I read deep linking. Thanks!
i saw something similar here: How do you authenticate a VueJS app with Azure AD?
but it did not work for me...
my problem is that after authenticating the user at login - i still needed to get the users security groups and that information was not received using the graph-api described in the above mentioned post
thank you for any help
it was something that took me a long time to figure out so im posting my findings here, hopfully this will help someone:
this was a hard one for me so im posting here - hopfully this will save some time to someone:
my problem was that i need not only to authenticate my vuejs app with azure-ad, but i need also to get the security groups to which the user is belonging to.
to achive this, this is what i done:
i used the vue-adal sample app mentioned above( you can find it in: https://github.com/survirtual/vue-adal ) - under sample folder.
but i still had to make some changes to make it behave the way i need. the problam was that after logging in with my user the sample app used windows.net graph api for retrieving user info with the token from the user authentication, so i had to change in main.js this:
const graphApiBase = `https://graph.windows.net`
const graphApiResource = '00000002-0000-0000-c000-000000000000'
to this:
const graphApiBase = `https://graph.microsoft.com/v1.0`
const graphApiResource = '00000003-0000-0000-c000-000000000000'
in addition, inside the return url component i had to change the axios query to get the security groups to which the user belongs to..so i changed this (in the home.vue file):
async getUserInfo () {
let res = await this.$graphApi.get(`me`, {
params: {
'api-version': 1.6
}
})
to this:
async getUserInfo () {
let res = await this.$graphApi.post(`/me/getMemberGroups`, {
securityEnabledOnly: true
})
console.log(res)
return res.data
}
and then the data that i received back from the api contained the security groups to which the user belongs to...
UPDATE 1
I removed return from code and now links work on IOS.
But on android I can't open any app. Any idea?
I am trying to open different apps from my app.
return Linking.openURL(“twitter://“);
return Linking.openURL(“instagram://“);
But it doesn’t work. I configured IOS by documentation. On android doesn’t work too. While...
return Linking.openURL(“tripadvisor://“);
Work just fine.
Any idea why I can’t open other apps.
This is code that I am using (open app if installed or open store with it but sometimes even store doesn't open) what I did wrong:
let appUrl = "instagram://";
Linking.canOpenURL(appUrl).then(supported => {
if (!supported) {
Alert.alert("",
"",
[
{text: "go to store", onPress: this.openStorePress},
{text: "cancel", onPress: () => { }, style: 'cancel'},
],
{ cancelable: false }
);
} else {
return Linking.openURL(appUrl);
}
}).catch(err => {
console.error(err);
});
Your issue is related to the content of the url, twitter:// means nothing for the Android Twitter app, so it will not open.
For example, the following code should work:
Linking.openURL('twitter://timeline')
or
Linking.openURL('instagram://user?username=apple')
You have to find the rights url schemes (documentations are not very clear about it) that may be different between iOS and Android.
Twitter: How can I open a Twitter tweet using the native Twitter app on iOS?
Instagram: https://www.instagram.com/developer/mobile-sharing/iphone-hooks/ (all do not work on Android)
misc: https://pureoxygenlabs.com/10-app-url-schemes-for-marketers/
You have to find the rights URL schemes. Have look at my code
Instagram
Linking.openURL('instagram://user?username=apple')
.catch(() => {
Linking.openURL('https://www.instagram.com/apple');
})
Twitter
Linking.openURL('twitter://user?screen_name=apple')
.catch(() => {
Linking.openURL('https://www.twitter.com/apple');
})
Facebook
Linking.openURL('fb://page/PAGE_ID');
Linking.openURL('http://instagram.com/_u/USER_NAME');
Linking.openURL('http://instagram.com/_p/PICTURE');
Your code looks pretty solid, here's an example of how I open twitter in my app.
const twitterUrlScheme = `twitter://user?screen_name=${twitterUsername}`;
Linking.canOpenURL(twitterUrlScheme)
.then((supported) =>
Linking.openURL(
supported
? twitterUrlScheme
: `https://www.twitter.com/${twitterUsername}`
)
)
.catch((err) => console.error('An error occurred', err));
I think perhaps your issue might be the return Linking.openUrl, I'm not sure you need the return in that statement. Does it work if you remove the return? Otherwise, it might help to move your Alert outside of the then-block from canOpenUrl.
I have used only url and it's working both iOS and android
Linking.openURL('https://www.facebook.com/');
You haven't completed the " fot twitter and instagram, I don't know whether you made the same mistake in app too, if yes, fixing that might solve it.
Try to use a package like:
https://github.com/react-native-community/react-native-share
You can try to use only some of it's functions or look into the native code from there and create some bridge functions in the native code and then export them to be used in JS code.
Note: you will have to use real devices for the tests.
I want to add to my application a button. So when is pressed, the phone's contacts application is opened, and the "Add contact" page is displayed with some data sent from the application.
How can I do that?
Update:
After several weeks of investigation, I still have no idea about how to do this.
I tried with this library: https://github.com/rt2zz/react-native-contacts
But it writes the contact directly to the contact list, without opening the contacts app.
Any help Will be super greatly appreciated
I am able to open native add to contact page using react-native-contacts.
link is here
Steps to do that
1. integrate react-native-contacts and make sure you follow all the steps and integrate correctly.
2. use below method to open it.
check openContactForm method of the library.
openContactPicker = () =>{
let number="1234567890"; //replace with any number
let newPerson = {
phoneNumbers: [{
label: "mobile",
number: number,
}],
};
Contacts.openContactForm(newPerson, (err) => {
if (err) console.warn(err) ;
// form is open
});
};
Note: contact we need to add should be a well-defined object that library recognizes. please check this
You Can Use wix-react-native-contacts npm package for your functionality.
refer this link
We're doing this in the web version of our react application and our native Android app so our setup and everything is working fine. I'm trying to implement sharing an action in react-native using react-native-fbsdk. I'm following the Android code because it looks the closest to the react-native-fbsdk code.
Should I be using ShareApi.share or something else?
I tried creating an instance of ShareOpenGraphContent to use with ShareApi.share, but there's no constructor.
I wish they would provide more thorough documentation :s
Based on the code my colleague used for the ShareApi on Android it seems like react-native-fbsdk is missing a few things related to sharing actions.
ShareOpenGraphContent isn't directly exported from react-native-fbsdk so this
import { ShareOpenGraphContent } from 'react-native-fbsdk';
Actually doesn't work. There must be some way to use the ShareApi in react-native-fbsdk to share an action...I'm just missing something.
Someone help...please.
Thanks!!
I figured it out! I had to manually create an instance of the ShareOpenGraphContent object which has 3 mandatory properties: contentType, action and previewPropertyName. The react-native-fbsdk doesn't currently have a constructor for this object type.
ShareApi.canShare isn't mandatory, but it checks to ensure you have the correct permissions before trying to share. This would allow you to get the user logged in before trying in case their token expired, or the user hasn't agreed to the needed permissions yet.
const ogAction = new ShareOpenGraphAction('<your_facebook_namespace>' + ':' + '<your_facebook_action>');
ogAction.putString('song', 'https://<url_to_your_song_app_etc>');
ogAction.putString('place', '<fbPlacePageID>'');
ogAction.putNumber('fb:explicitly_shared', 1); // Normally this is a boolean, but putNumber with a value of 1 works
// Manually create an instance of ShareOpenGraphContent (no constructor in the API)
const ogContent = {
contentType: 'open-graph',
action: ogAction,
previewPropertyName: 'song',
};
ShareApi.canShare(ogContent).then((canShare) => {
if (canShare)
return ShareApi.share(ogContent, '/me');
}).then(
function(result) {
// Shared successfully
},
function(error) {
// Failed to share
}
);