Develop Reference

Android Intent to open instagram posting view

How to open Instagram app in posting mode (so the same view we get after clicking plus at home screen within Instagram app)? I've seen android apps which are doing that, however wasn't able to find information explaing how to achieve it.
That code works for opening the home view:
Uri uri = Uri.parse("http://instagram.com/");
Intent likeIng = new Intent(Intent.ACTION_VIEW, uri);
likeIng.setPackage("com.instagram.android");
startActivity(likeIng);
Perhaps there is some other uri which could work?

After a bit of research it turned out that deep linking with 'instagram://share' opens the desired view.

You can do that only when you have a media to share on Instagram:
try {
String type = "image/*";
String filename = "/myPhoto.jpg";
String mediaPath = Environment.getExternalStorageDirectory() + filename;
Intent share = new Intent(Intent.ACTION_SEND);
share.setPackage("com.instagram.android");
share.setType(type);
File media = new File(mediaPath);
Uri uri = null;
share.putExtra(Intent.EXTRA_STREAM, uri);
startActivity(share);
} catch (Exception e) {
e.printStackTrace();
}
You can create a fake media though to be able to use this method.

Just to expand on #Paavo answer to show the answer in code.
Uri uri = Uri.parse("instagram://share");
Intent intent = new Intent(Intent.ACTION_VIEW, uri);
intent.setPackage("com.instagram.android");
//Check that the user can open the intent, meaning they have the Instagram app installed
if (getPackageManager().resolveActivity(intent, 0) != null)
startActivity(intent);
else //User doesn't have app installed, handle however you want
throw new Resources.NotFoundException();

intent share facebook - Android

I want to share an image from android activity to facebook messenger app which is already opened in my device and reside in recent tasks of the android device.
I am using the below code to share the image but it is opening a new activity in messenger 'share seperately' but I want to share it to the already opened chat.
String mimeType = "image/*";
Intent intent = new Intent(Intent.ACTION_SEND);
intent.setPackage("com.facebook.orca");
intent.setType(mimeType);
intent.putExtra(Intent.EXTRA_STREAM, contentUri);
intent.putExtra(EXTRA_PROTOCOL_VERSION, PROTOCOL_VERSION);
intent.putExtra(EXTRA_APP_ID, YOUR_APP_ID);
activity.startActivityForResult(shareIntent, SHARE_TO_MESSENGER_REQUEST_CODE);

Facebook has restricted the text sharing over the post. It will either support a valid URL link sharing or Image Sharing. So I've posted the code which is referenced from https://stackoverflow.com/a/22994833/8331006
private void shareAppLinkViaFacebook(String urlToShare) {
try {
Intent intent1 = new Intent();
intent1.setClassName("com.facebook.katana", "com.facebook.katana.activity.composer.ImplicitShareIntentHandler");
intent1.setAction("android.intent.action.SEND");
intent1.setType("text/plain");
intent1.putExtra("android.intent.extra.TEXT", urlToShare);
startActivity(intent1);
} catch (Exception e) {
// If we failed (not native FB app installed), try share through SEND
String sharerUrl = "https://www.facebook.com/sharer/sharer.php?u=" + urlToShare;
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(sharerUrl));
startActivity(intent);
}
}

Facebook Link doesnt work android

Hello I just want to call Facebook app with the link below (on my android project):
String url_facebook_prixo = "https://www.facebook.com/pages/Prixo/468580313168290";
I tried this :
Uri uri = Uri.parse("facebook://facebook.com/page{468580313168290}");
Intent viewIntent = new Intent(Intent.ACTION_VIEW, uri);
startActivity(viewIntent);
I tried with others link but its only displaying my wall..
Someone?

You should use the id to open the page.
Intent facebookIntent = new Intent(
Intent.ACTION_VIEW,
Uri.parse("fb://profile/468580313168290"));
startActivity(facebookIntent);
I would advice you to encapsulate this in a try catch. Inside the catch open a normal browser intent

Open page in Twitter app from other app - Android

I was looking for some way to launch Twitter app and open a specified page from my application, without webview.
I found the solution for Facebook here:
Opening facebook app on specified profile page
I need something similar.
[EDIT]
I've just found a solution:
try {
Intent intent = new Intent(Intent.ACTION_VIEW,
Uri.parse("twitter://user?screen_name=[user_name]"));
startActivity(intent);
} catch (ActivityNotFoundException e) {
startActivity(new Intent(Intent.ACTION_VIEW,
Uri.parse("https://twitter.com/#!/[user_name]")));
}

Based on fg.radigales answer, this is what I used to launch the app if possible, but fall back to the browser otherwise:
Intent intent = null;
try {
// get the Twitter app if possible
this.getPackageManager().getPackageInfo("com.twitter.android", 0);
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("twitter://user?user_id=USERID"));
intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
} catch (Exception e) {
// no Twitter app, revert to browser
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://twitter.com/PROFILENAME"));
}
this.startActivity(intent);
UPDATE
Added intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK); to fix an issue where twitter was opening inside my app instead of as a new activity.

This worked for me: twitter://user?user_id=id_num

Open page on Twitter app from other app using Android in 2 Steps:
1.Just paste the below code (on button click or anywhere you need)
Intent intent = null;
try{
// Get Twitter app
this.getPackageManager().getPackageInfo("com.twitter.android", 0);
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("twitter://user?user_id=USER_ID"));
intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
} catch () {
// If no Twitter app found, open on browser
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://twitter.com/USERNAME"));
}
2.intent = new Intent(Intent.ACTION_VIEW, Uri.parse("twitter://user?user_id=USER_ID"));
To get USER_ID just write username https://tweeterid.com/ and get Twitter User ID in there
Reference: https://solutionspirit.com/open-page-twitter-application-android/

My answer builds on top of the widely-accepted answers from fg.radigales and Harry.
If the user has Twitter installed but disabled (for example by using App Quarantine), this method will not work. The intent for the Twitter app will be selected but it will not be able to process it as it is disabled.
Instead of:
getPackageManager().getPackageInfo("com.twitter.android", 0);
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("twitter://user?user_id=2343965036"));
You can use the following to decide what to do:
PackageInfo info = getPackageManager().getPackageInfo("com.twitter.android", 0);
if(info.applicationInfo.enabled)
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("twitter://user?user_id=2343965036"));
else
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://twitter.com/wrkoutapp"));

Just try this code snippet. It will help you.
//Checking If the app is installed, according to the package name
Intent intent = new Intent();
intent.setType("text/plain");
intent.setAction(Intent.ACTION_SEND);
final PackageManager packageManager = getPackageManager();
List<ResolveInfo> list = packageManager.queryIntentActivities(intent, PackageManager.MATCH_DEFAULT_ONLY);
for (ResolveInfo resolveInfo : list)
{
String packageName = resolveInfo.activityInfo.packageName;
//In case that the app is installed, lunch it.
if (packageName != null && packageName.equals("com.twitter.android"))
{
try
{
String formattedTwitterAddress = "twitter://user/" ;
Intent browseTwitter = new Intent(Intent.ACTION_VIEW, Uri.parse(formattedTwitterAddress));
long twitterId = <Here is the place for the twitter id>
browseTwitter.putExtra("user_id", twitterId);
startActivity(browseTwitter);
return;
}
catch (Exception e)
{
}
}
}
//If it gets here it means that the twitter app is not installed. Therefor, lunch the browser.
try
{
String twitterName = <Put the twitter name here>
String formattedTwitterAddress = "http://twitter.com/" + twitterName;
Intent browseTwitter = new Intent(Intent.ACTION_VIEW, Uri.parse(formattedTwitterAddress));
startActivity(browseTwitter);
}
catch (Exception e)
{
}

For me this did the trick it opens Twitter app if you have it or goes to web browser:
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://twitter.com/"+"USERID"));
startActivity(intent);

try {
Intent intent = new Intent(Intent.ACTION_VIEW,
Uri.parse("twitter://user?screen_name=[user_name]"));
startActivity(intent);
} catch (Exception e) {
startActivity(new Intent(Intent.ACTION_VIEW,
Uri.parse("https://twitter.com/#!/[user_name]")));
}
This answer was posted as an edit to the question Open page in Twitter app from other app - Android by the OP jbc25 under CC BY-SA 3.0.

Open a facebook page from android app?

How can I start an intent to open a Facebook application on a phone and navigate to the prefered page in Facebook?
I tried:
Intent intent = new Intent(Intent.ACTION_VIEW);
intent.setClassName("com.facebook.katana", "com.facebook.katana.ProfileTabHostActivity");
intent.putExtra("extra_user_id", "123456789l");
this.startActivity(intent);
Well, whatever I write to "1234567891", it is always navigating to my page. Always to me and not else.
How could I do this?

Here is the best and simple way to do it.
Just follow the code
public final void Facebook() {
final String urlFb = "fb://page/"+yourpageid;
Intent intent = new Intent(Intent.ACTION_VIEW);
intent.setData(Uri.parse(urlFb));
// If Facebook application is installed, use that else launch a browser
final PackageManager packageManager = getPackageManager();
List<ResolveInfo> list =
packageManager.queryIntentActivities(intent,
PackageManager.MATCH_DEFAULT_ONLY);
if (list.size() == 0) {
final String urlBrowser = "https://www.facebook.com/pages/"+pageid;
intent.setData(Uri.parse(urlBrowser));
}
startActivity(intent);
}

I had the exactly same problem, sent the user id but for some reason, my profile always opened instead of the friend's profile.
The problem is that if you pass the String of the Long object that represents the Facebook UID, or even a long primitive type, the intent won't be able to read it later. You need to pass a real Long.
So the complete code is:
Intent intent = new Intent(Intent.ACTION_VIEW);
intent.setClassName("com.facebook.katana", "com.facebook.katana.ProfileTabHostActivity");
Long uid = new Long("123456789");
intent.putExtra("extra_user_id", uid);
startActivity(intent);
Ok enjoy and hope this helps :-)
Maxim

try this code :
String facebookUrl = "https://www.facebook.com/<id_here>";
try {
int versionCode = getPackageManager().getPackageInfo("com.facebook.katana", 0).versionCode;
if (versionCode >= 3002850) {
Uri uri = Uri.parse("fb://facewebmodal/f?href=" + facebookUrl);
startActivity(new Intent(Intent.ACTION_VIEW, uri));
} else {
Uri uri = Uri.parse("fb://page/<id_here>");
startActivity(new Intent(Intent.ACTION_VIEW, uri));
}
} catch (PackageManager.NameNotFoundException e) {
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(facebookUrl)));
}

This solution won't work any more. The new version of the facebook app doesn't support anymore those kind of intents. See here the bug report
The new solution is to use the iPhone scheme mechanism (Yes, facebook decided to support the iPhone mechanism in Android instead of the implicit intent mechanism of Android).
So in order to open the facebook app with a user profile all you need to do is:
String facebookScheme = "fb://profile/" + facebookId;
Intent facebookIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(facebookScheme));
startActivity(facebookIntent);
If you are looking for other actions you can use the following page for all available actions (/you have to test it though, since I didn't find an official publication of facebook about this)