Opening facebook profile page in facebook native app from my android app - android

I'm trying to open a specific profile page from my android app, in the native facebook app. I've seen many answers about this topic, but none of them worked for me currently (maybe because some new versions)
Right now, I'm trying to open it like that:
public static Intent getFacebookIntent(PackageManager pm, String url ,String fbID) {
Intent intent;
try {
int versionCode = pm.getPackageInfo("com.facebook.katana", 0).versionCode;
boolean activated = pm.getApplicationInfo("com.facebook.katana", 0).enabled;
if(activated){
if ((versionCode >= 3002850)) {
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("fb://facewebmodal/f?href=" + url));
return intent;
} else {
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("fb://page/" + fbID));
return intent;
}
}else{
intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
return intent;
}
} catch (PackageManager.NameNotFoundException e) {
intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
return intent;
}
}
where fbID its the user id, and url is the "link" from user graph request.
I've tried to change the url, to a specific url like
"https://www.facebook.com/some_user_name" but not luck.
the result of that code, is that the facebook app opens and crashes.
I do able to open it with webpage link.
Thanks

Related

Why is "complete action using" popping up for this intent?

I just linked some social networks to my app as a preliminary test and using the same kind of code, the results I am getting are different for Facebook, Instagram and Twitter intents.
When I click Facebook or Twitter, it opens the app automatically when it's installed and uses browser when it's not. However, that's not the case with Instagram. The complete action using dialog pops up and that's not something I want to happen.
protected void LaunchInstagram() {
String InstagramUsername = "USERNAME";
String LaunchInstagram = "http://instagram.com/_u/" + InstagramUsername;
String InstagramURL = "https://instagram.com/" + InstagramUsername;
try {
this.getPackageManager().getPackageInfo("com.instagram.android", 0);
Uri Uri = Uri.parse(LaunchInstagram);
startActivity(new Intent(Intent.ACTION_VIEW, Uri));
} catch (PackageManager.NameNotFoundException InstagramAppNotFoundOpenBrowser) {
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(InstagramURL)));
}
}
You are using
String InstagramURL = "https://instagram.com/" + InstagramUsername;
So every Intent you'll try to launch, will open a chooser because this is a URL.
than every App that supports URL, like browsers and such, will try to handle that.
EDIT 1:
Try it this way:
public static boolean openApp(Context context, String packageName) {
PackageManager manager = context.getPackageManager();
try {
Intent i = manager.getLaunchIntentForPackage(packageName);
if (i == null) {
return false;
//throw new PackageManager.NameNotFoundException();
}
i.addCategory(Intent.CATEGORY_LAUNCHER);
context.startActivity(i);
return true;
} catch (PackageManager.NameNotFoundException e) {
return false;
}
}
I'm not sure if that will help you to open a specific user page on the App thou.
EDIT 2:
Try it this way:
Uri uri = Uri.parse("http://instagram.com/_u/USERNAME");
Intent insta = new Intent(Intent.ACTION_VIEW, uri);
insta.setPackage("com.instagram.android");
startActivity(insta);

Open facebook page from android app (in facebook version > v11)

I used to open my facebook page from my app using the below code, but this does not work anymore starting facebook v11.0.0.11.23 released on June 21, 2014, any idea how to open the page in the new facebook app?
To note that it opens now the facebook app but without opening the specified page, it used to work just fine before the latest update
public void openFacebookPage() {
Intent intent = null;
try {
context.getPackageManager().getPackageInfo("com.facebook.katana", 0);
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("fb://page/<id here>"));
//tried this also
//intent = new Intent(Intent.ACTION_VIEW, Uri.parse("fb://profile/<id here>"));
} catch (Exception e) {
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://www.facebook.com/<name here>"));
}
context.startActivity(intent);
}
In Facebook version 11.0.0.11.23 (3002850) fb://profile/ and fb://page/ are no longer supported. I decompiled the Facebook app and was able to come up with the following solution:
String facebookUrl = "https://www.facebook.com/JRummyApps";
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 {
// open the Facebook app using the old method (fb://profile/id or fb://page/id)
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("fb://page/336227679757310")));
}
} catch (PackageManager.NameNotFoundException e) {
// Facebook is not installed. Open the browser
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(facebookUrl)));
}
Edit: It has been some time and it looks like fb://profile and fb://page are no longer supported. Below is the method I have been using in production:
/**
* Intent to open the official Facebook app. If the Facebook app is not installed then the
* default web browser will be used.</p>
*
* Example usage:</p>
* <code>newFacebookIntent(context.getPackageManager(), "https://www.facebook.com/JRummyApps");</code></p>
*
* #param pm
* Instance of the {#link PackageManager}.
* #param url
* The full URL to the Facebook page or profile.
* #return An intent that will open the Facebook page/profile.
*/
public static Intent newFacebookIntent(PackageManager pm, String url) {
Uri uri;
try {
pm.getPackageInfo("com.facebook.katana", 0);
// http://stackoverflow.com/a/24547437/1048340
uri = Uri.parse("fb://facewebmodal/f?href=" + url);
} catch (PackageManager.NameNotFoundException e) {
uri = Uri.parse(url);
}
return new Intent(Intent.ACTION_VIEW, uri);
}

Android open specific facebook app profile page at other android app redirect to newsfeed page when facebook app is closed

I am trying to open certain Facebook profile page in my android app with specific user ID.
My code opens facebook app if it is installed otherwise open webbrowser.
It works fine except when facebook app is installed and it is closed.
In that situation it just opens newsfeed page instead of the profile page. When the facebook app is open at background, it successfully redirect to desired profile page. How can I solve this ? Also is there an official facebook document which describes about the way to access facebook app URI ?
try{
this.getPackageManager()
.getPackageInfo("com.facebook.katana", 0); //Checks if FB is even installed.
final String facebookScheme = String.format("fb://profile/%s", user2FbID);
final Intent facebookIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(facebookScheme));
facebookIntent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK | Intent.FLAG_ACTIVITY_NO_HISTORY
| Intent.FLAG_ACTIVITY_CLEAR_WHEN_TASK_RESET);
startActivity(facebookIntent);
}catch (Exception e) {
String facebookScheme = "https://m.facebook.com/" + user2FbID;
Intent facebookIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(facebookScheme));
startActivity(facebookIntent);
}
Try this, Works for me
try
{
Intent followIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("fb://profile/<your profile_id>"));
startActivity(followIntent);
final Handler handler = new Handler();
handler.postDelayed(new Runnable()
{
#Override
public void run() {
Intent followIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("fb://profile/<your profile_id>"));
startActivity(followIntent);
}
}, 1000 * 2);
}
catch (Exception e)
{
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("https://www.facebook.com/<user_name>")));
String errorMessage = (e.getMessage()==null)?"Message is empty":e.getMessage();
Log.e("Unlock_ScreenActivity:FacebookAppNotFound" ,errorMessage);
}
This was happening with me too. Updated the Facebook app and it's solved now, seems it was something to do in with Facebook. It's working properly now.
Here's the code I'm using:
public static Intent getOpenFacebookIntent(Context context) {
try{
// open in Facebook app
context.getPackageManager().getPackageInfo("com.facebook.katana", 0);
return new Intent(Intent.ACTION_VIEW, Uri.parse("fb://page/<profile_id>"));
} catch (Exception e) {
// open in browser
return new Intent(Intent.ACTION_VIEW, Uri.parse("https://www.facebook.com/<profile_id>"));
}
}

Open Instagram User Profile

I managed to open the Twitter and Facebook user profile from my app. But I can not find any references to Instagram.
Is There a way to open Instagram in order to show a user profile like in twitter or facebook?
For instance, in order to get the Intent to launch the twitter application I do:
public Intent getOpenTwitterIntent(Context context) {
try {
return new Intent(Intent.ACTION_VIEW,
Uri.parse("twitter://user?screen_name="
.concat(twitterUsername)));
} catch (Exception e) {
return new Intent(
Intent.ACTION_VIEW,
Uri.parse("https://twitter.com/#!/".concat(twitterUsername)));
}
}
How can I achive something similar with Instagram?
Thanks in advance.
Here is a method to get the Intent to open the Instagram app to the user's profile page:
/**
* Intent to open the official Instagram app to the user's profile. If the Instagram app is not
* installed then the Web Browser will be used.</p>
*
* Example usage:</p> {#code newInstagramProfileIntent(context.getPackageManager(),
* "http://instagram.com/jaredrummler");}</p>
*
* #param pm
* The {#link PackageManager}. You can find this class through
* {#link Context#getPackageManager()}.
* #param url
* The URL to the user's Instagram profile.
* #return The intent to open the Instagram app to the user's profile.
*/
public static Intent newInstagramProfileIntent(PackageManager pm, String url) {
final Intent intent = new Intent(Intent.ACTION_VIEW);
try {
if (pm.getPackageInfo("com.instagram.android", 0) != null) {
if (url.endsWith("/")) {
url = url.substring(0, url.length() - 1);
}
final String username = url.substring(url.lastIndexOf("/") + 1);
// http://stackoverflow.com/questions/21505941/intent-to-open-instagram-user-profile-on-android
intent.setData(Uri.parse("http://instagram.com/_u/" + username));
intent.setPackage("com.instagram.android");
return intent;
}
} catch (NameNotFoundException ignored) {
}
intent.setData(Uri.parse(url));
return intent;
}
So far I couldn't find a way to show user's profile directly.. but there is a way around..
Found this solution from Instagram Manifest from GitHub
Intent iIntent = getPackageManager().getLaunchIntentForPackage("com.instagram.android");
iIntent.setComponent(new ComponentName( "com.instagram.android", "com.instagram.android.activity.UrlHandlerActivity"));
iIntent.setData( Uri.parse( "http://instagram.com/p/gjfLqSBQTJ/") );
startActivity(iIntent);
This will open particular image post by the user in the app. From the page app user can open the profile with the link inside.
If you want to open recent post or something , you can use Instagram API
This is not we want exactly , but the better option now... i guess :)
After a brief search and trying what we all know "WON'T WORK" I got this to work
Uri uri = Uri.parse("http://instagram.com/mzcoco2you");
Intent intent = new Intent(Intent.ACTION_VIEW, uri);
startActivity(intent);
This should start your default browser and ... there you go. The answer is "instagram.com" + "/UserName".
To open directly instagram app to a user profile :
String scheme = "http://instagram.com/_u/USER";
String path = "https://instagram.com/USER";
String nomPackageInfo ="com.instagram.android";
try {
activite.getPackageManager().getPackageInfo(nomPackageInfo, 0);
intentAiguilleur = new Intent(Intent.ACTION_VIEW, Uri.parse(scheme));
} catch (Exception e) {
intentAiguilleur = new Intent(Intent.ACTION_VIEW, Uri.parse(path));
}
activite.startActivity(intentAiguilleur);
// Use this link to open directly a picture
String scheme = "http://instagram.com/_p/PICTURE";
try {
Intent intent = new Intent(Intent.ACTION_VIEW);
intent.setData(Uri.parse("http://instagram.com/_u/" + "username"));
intent.setPackage("com.instagram.android");
startActivity(intent);
}
catch (android.content.ActivityNotFoundException anfe)
{
startActivity(new Intent(Intent.ACTION_VIEW,
Uri.parse("https://www.instagram.com/" + username)));
}
Unfortunately, at this time you cannot open the Instagram app to go directly to a user profile. You can however open a certain photo in the Instagram app.
The following code that I have provided will open a certain photo inside of the Instagram app. If no Instagram app is installed, it will open a user profile inside of the browser.
public void openInstagram (View view) {
Intent intent = null;
String pictureId = "p/0vtNxgqHx9";
String pageName = "d4v3_r34d";
try {intent = getPackageManager().getLaunchIntentForPackage("com.instagram.android");
intent.setComponent(new ComponentName("com.instagram.android", "com.instagram.android.activity.UrlHandlerActivity"));
intent.setData(Uri.parse("http://instagram.com/" + pictureId));
} catch (Exception e) {
intent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://instagram.com/" + pageName));
}
this.startActivity(intent);
I made it very easy for this code to be edited for your own purposes. Set the "String pictureId" to the id of the picture that you want to display and set the "String pageName" to the user name of your Instagram account.
The user name part is obvious but if you need help to get your picture id look at this picture.
Check the code below:
Intent instaintent = getPackageManager().getLaunchIntentForPackage("com.instagram.android");
instaintent.setData(Uri.parse("https://www.instagram.com/insta_save_new/"));
startActivity(instaintent);

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.

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