Develop Reference

String uri to open a page in facebook lite app

When I wanna open a page on normal facebook app (calling it through my app) I use the follow string to make the Uri.parse().
"fb://page/[pageId]", and it works normally
I do this:
String uriStr = "fb://page/[pageId]";
Uri uri = Uri.parse(uriStr);
startActivity(new Intent(Intent.ACTION_VIEW, uri));
However when I try open it in facebook lite, the same doesn't work, and my app is closed by some error. :( Exist another way? What is the correct string to open the same page in facebook lite? Anyone knows? Thx
EDIT
LogCat error:
android.content.ActivityNotFoundException: No Activity found to handle Intent

int versionCode = getPackageManager().getPackageInfo("com.facebook.lite", 0).versionCode;
if (versionCode >= 3002850) { //newer versions osf fb app
link = "fb://facewebmodal/f?href=" + FACEBOOK_URL;
} else { //older versions of fb app
link = "fb://page/" + FACEBOOK_URL;
}}

Look like this
String uri = "facebook://facebook.com/inbox";
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(uri));
startActivity(intent);
or for special activity
Intent intent = new Intent("android.intent.category.LAUNCHER");
intent.setClassName("com.facebook.katana", "com.facebook.katana.LoginActivity");
startActivity(intent);
hope it help.

uri = Uri.parse("fb://facewebmodal/f?href=" + url);
new Intent(Intent.ACTION_VIEW, uri);

Android BrowserActivity on Kindle HD

I have an Android app that creates a local HTML file & then displays this to the user in a browser. I have had issues with BrowserActivity not working on different devices, subject to whatever browser is installed. My current code does the following -
public void displayStats()
{
String file = produceStats();
Uri uri = Uri.parse("file://" + file);
// chrome ??
Intent intent1 = new Intent(Intent.ACTION_VIEW);
intent1.setDataAndType(uri, "multipart/related");
// default "Internet" browser
Intent intent2 = new Intent(Intent.ACTION_VIEW, uri);
intent2.setDataAndType(uri, "text/html");
intent2.setClassName("com.android.browser", "com.android.browser.BrowserActivity");
// any other browser (FireFox) ??
Intent intent3 = new Intent(Intent.ACTION_VIEW);
intent3.setDataAndType(uri, "text/html");
intent3.addCategory(Intent.CATEGORY_BROWSABLE);
PackageManager packageManager = getPackageManager();
List<ResolveInfo> activities1 = packageManager.queryIntentActivities(intent1, 0);
List<ResolveInfo> activities2 = packageManager.queryIntentActivities(intent2, 0);
List<ResolveInfo> activities3 = packageManager.queryIntentActivities(intent3, 0);
boolean isIntentSafe1 = activities1.size() > 0;
boolean isIntentSafe2 = activities2.size() > 0;
boolean isIntentSafe3 = activities3.size() > 0;
List<Intent> targetedShareIntents = new ArrayList<Intent>();
if (isIntentSafe1)
{
unpackResolvedIntents(uri, "multipart/related", activities1, targetedShareIntents);
}
if (isIntentSafe2) {
unpackResolvedIntents(uri, "text/html", activities2, targetedShareIntents);
}
if (isIntentSafe3) {
unpackResolvedIntents(uri, "text/html", activities3, targetedShareIntents);
}
if (targetedShareIntents.isEmpty()) {
// go to market to install app ????
Toast.makeText(plink.this, "Please install BROWSER to complete (Chrome)", Toast.LENGTH_LONG).show();
Intent goToMarket = new Intent(Intent.ACTION_VIEW)
.setData(Uri.parse("market://details?id=com.android.chrome"));
startActivity(goToMarket);
} else if (targetedShareIntents.size() == 1) {
startActivity(targetedShareIntents.remove(0));
} else {
Intent chooserIntent = Intent.createChooser(targetedShareIntents.remove(0), "Select viewer");
Intent[] extraIntents = new Intent[targetedShareIntents.size()];
for (int i = 0; i < targetedShareIntents.size(); i++) {extraIntents[i] = targetedShareIntents.get(i);}
chooserIntent.putExtra(Intent.EXTRA_INITIAL_INTENTS, extraIntents);
startActivity(chooserIntent);
}
}
The produceStats() call returns the path to the file, then the rest of this function handles various different browser, and if more than one available it offers the user a Chooser.
My problem is that one user has reported the app crashing when he runs this on a Kindle HD device with the SILK browser. His stack dump is thus -
26 Jan 2014 16:26:09 GMT:Uncaught exception in java.lang.Thread:main(Unable to find explicit activity class {com.android.browser/com.android.browser.BrowserActivity}; have you declared this activity in your AndroidManifest.xml?)
android.content.ActivityNotFoundException: Unable to find explicit activity class {com.android.browser/com.android.browser.BrowserActivity}; have you declared this activity in your AndroidManifest.xml?
at android.app.Instrumentation.checkStartActivityResult(Instrumentation.java:1624)
at android.app.Instrumentation.execStartActivity(Instrumentation.java:1418)
My question is - how do I launch on a Kindle to display the file in SILK ?
Thank you

Slik on a Kindle Fire HD does not seem to support the file:// scheme for HTML, based upon an examination of its manifest. It only appears to support http://, https://, and inline://. I cannot explain the specific crash that you are encountering, as I do not see any sign of the AOSP browser app, and so I do not know why PackageManager would report otherwise.

final Uri uri = Uri.parse(filePath);
Intent browserIntent = new Intent(Intent.ACTION_VIEW, uri);
browserIntent.addCategory(Intent.CATEGORY_BROWSABLE);
context.startActivity(browserIntent);

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)

Android launch browser without specifying a URL

How would one go about launching the browser from an activity without specifying a url. I would like to open the browser so the user can continue browsing without changing the page they were on?
SOLUTION:
Answer below was correct and worked, but to be more specific for future readers, here is the working code:
Intent i = new Intent();
i.setAction(Intent.ACTION_MAIN);
i.addCategory(Intent.CATEGORY_LAUNCHER);
i.setAction("com.android.browser");
ComponentName comp = new ComponentName("com.android.browser", "com.android.browser.BrowserActivity");
i.setComponent(comp);
i.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
startActivity(i);
Thanks!

Use Intent#setComponent() to set the Browser's package and class name. Then start the activity.

In case it (the ComponentName("com.android.browser", "com.android.browser.BrowserActivity")) changes in the future, you may try something like the following code:
public static ComponentName getDefaultBrowserComponent(Context context) {
Intent i = new Intent()
.setAction(Intent.ACTION_VIEW)
.setData(new Uri.Builder()
.scheme("http")
.authority("x.y.z")
.appendQueryParameter("q", "x")
.build()
);
PackageManager pm = context.getPackageManager();
ResolveInfo default_ri = pm.resolveActivity(i, 0); // may be a chooser
ResolveInfo browser_ri = null;
List<ResolveInfo> rList = pm.queryIntentActivities(i, 0);
for (ResolveInfo ri : rList) {
if (ri.activityInfo.packageName.equals(default_ri.activityInfo.packageName)
&& ri.activityInfo.name.equals(default_ri.activityInfo.name)
) {
return ri2cn(default_ri);
} else if ("com.android.browser".equals(ri.activityInfo.packageName)) {
browser_ri = ri;
}
}
if (browser_ri != null) {
return ri2cn(browser_ri);
} else if (rList.size() > 0) {
return ri2cn(rList.get(0));
} else if (default_ri == null) {
return null;
} else {
return ri2cn(default_ri);
}
}
private static ComponentName ri2cn(ResolveInfo ri) {
return new ComponentName(ri.activityInfo.packageName, ri.activityInfo.name);
}
Basically, here I construct an intent to view a dummy http page, get the list of activities that may handle the intent, compare it to the default handler returned by resolveActivity() and return something. I do not need to check if there's a launcher MAIN action (my code uses the VIEW action), but you probably should.

this answer may help. from How to open the default android browser without specifying an URL?
PackageManager pm = getPackageManager();
Intent queryIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.google.com"));
ActivityInfo af = queryIntent.resolveActivityInfo(pm, 0);
Intent launchIntent = new Intent(Intent.ACTION_MAIN);
launchIntent.setClassName(af.packageName, af.name);
startActivity(launchIntent);