Develop Reference

Android testing - change system value for testing logical branch

I've got a method that tries to start a browser with a specific URL, or if the device does not have a browser installed, instead starts an Activity with a WebView and loads the same URL. Basically this:
try {
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
getActivity().startActivity(intent);
} catch (ActivityNotFoundException e) {
Intent intent = new Intent(getActivity(), MyWebView.class);
intent.putExtra(EXTRA_URL, url);
getActivity().startActivity(intent);
}
}
How can I test both branches? Is it possible to unregister the Browser activity from the device, temporarily and programmatically?
I had considered a Shadow Activity that just throws an ActivityNotFoundException when that particular Intent was passed to it (i'd check the data URI), but AFAIK there's no way to turn that Shadow on for one test and off for the other.
TYIA

This is really over engineering. The odds that the user doesn't have a web browser on the device is 0. It isn't worth worrying about.
If you're worried about the url being invalid and unparsable, its unlikely this would fix that problem, as it would be unparsable in your web view activity as well. If you really want to be paranoid I'd catch it and toast an error.
As for testing it- refactor to make it testable. Write this as a function that takes a URL as a parameter. For the bad version, use a URL like thisisnotavalidscheme://example.com. That will fail as nobody is going to be registered for that scheme.

Intent to open a chat with a specific user on snapchat app

I'm trying to find if there is any app schema,
to open the Snapchat app (via Intent) with a specific userID that I want to chat with?
BTW, to find the userID:

This the only thing that works for me. Unfortunately, it adds the extra step of making the user choose the browser or Snapchat app.
Intent nativeAppIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://snapchat.com/add/" + snapchatId));
startActivity(nativeAppIntent);
Oddly enough, the URL scheme snapchat://add/" + snapchatId works on iOS but not on Android (it opens the Android app, but does not pop up the user).
EDIT: Add intent.setPackage("com.snapchat.android"); and it will open the app without the chooser. But adding this means you will need to surround everything with a try/catch to prevent a crash.
try {
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse("https://snapchat.com/add/" + snapchatId));
intent.setPackage("com.snapchat.android");
startActivity(intent);
} catch (Exception e) {
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("https://snapchat.com/add/" + snapchatId)));
}

Launching Android Netflix App And Passing Video Id

In the app I am working on I want to support Netfilx streaming. I intend on doing this by simply starting Netflix and passing a specific URI so it plays a specific video when started. Simple right? Well, the issue is I'm not sure how to pass the video id info in the Intent I use to start the Activity.
I've read the post here , but am unsure where to use this. I used Intent.setData() since it accepts a URI, but to no avail.
Here is what I have been doing (I am hard coding the movie data, this is just for testing purposes) :
// the Netflix intent
Intent intent = getPackageManager().getLaunchIntentForPackage("com.netflix.mediaclient");
//the uri
Uri uri = Uri.parse("http://movies.netflix.com/WiPlayer?movieid=70266228&trkid=13462049&ctx=0%2C1%2Ce2bd7b74-6743-4d5e-864f-1cc2568ba0da-61921755");
intent.setData(uri);
//launches but does not go to the video
startActivity(intent);
I've also tried using the URI protocol in the link above like so:
Uri uri = Uri.parse("nflx://movies.netflix.com/WiPlayer?movieid=70266228&trkid=13462049&ctx=0%2C1%2Ce2bd7b74-6743-4d5e-864f-1cc2568ba0da-61921755");
but still am not seeing the video play.
I feel like I am missing something simple here, although I have had very little luck Googling for this, I could find next to nothing about starting the Netflix Android app from another application. The Netflix developer resources don't have any info on this.
Does anyone have any suggestions on how I can do this or where I should be looking for documentation on this? Any help would be appreciated. Thanks much!

Just some Android Apps intent names to help anyone.
Format: appName, packageName, className
Skype: com.skype.raider, com.skype.raider.Main
Netflix: com.netflix.mediaclient, com.netflix.mediaclient.ui.launch.UIWebViewActivity
ESexplorer: com.estrongs.android.pop, com.estrongs.android.pop.view.FileExplorerActivity
Youtube: com.google.android.youtube,com.google.android.youtube.HomeActivity
Chrome: com.android.chrome,com.google.android.apps.chrome.Main
VLC: org.videolan.vlc, org.videolan.vlc.gui.MainActivity
MBOXSettings: com.mbx.settingsmbox, com.mbx.settingsmbox.SettingsMboxActivity

I've managed to do this. With the following code. First you need the movieId (or videoId) for netflix, then compose the URL to watch.
String netFlixId = "43598743"; // <== isn't a real movie id
String watchUrl = "http://www.netflix.com/watch/"+netFlixId;
Then with this intent
try {
Intent intent = new Intent(Intent.ACTION_VIEW);
intent.setClassName("com.netflix.mediaclient", "com.netflix.mediaclient.ui.launch.UIWebViewActivity");
intent.setData(Uri.parse(watchUrl));
startActivity(intent);
}
catch(Exception e)
{
// netflix app isn't installed, send to website.
Intent intent = new Intent(Intent.ACTION_VIEW);
intent.setData(Uri.parse(item.url));
startActivity(intent);
}
I haven't tried with TV Shows yet. But this works beautifully. If you want to send to the profile page of the movie.. Send it to a url formatted this way
http://www.netflix.com/title/324982
I also found out how to search for a title.
try {
Intent intent = new Intent(Intent.ACTION_SEARCH);
intent.setClassName("com.netflix.mediaclient", "com.netflix.mediaclient.ui.search.SearchActivity");
intent.putExtra("query", item.label);
startActivity(intent);
}
catch(Exception e)
{
Toast.makeText(this, "Please install the NetFlix App!", Toast.LENGTH_SHORT).show();
}

Here is how to start a movie from an ADB command for com.netflix.mediaclient
adb shell am start -n com.netflix.mediaclient/.ui.launch.UIWebViewActivity -a android.intent.action.VIEW -d http://www.netflix.com/watch/60000724
I still can't find a way to do it for com.netflix.ninja (Netflix for Android TV)

Share via Google+..dialog complete action with shown

hi everybody i've a little problem..i'm trying to share a text using google plus app, but the lastest version give me a problem.
when i call with an intent "com.google.android.apps.plus", android display me this dialog
so if i choose the first one, text will be shared correctly, but the second one do nothing.
if the app is not intalled i redirect the user to the market, at g+ download page (this work fine)
if (v == plus) {
social(2);
targetedShareIntent.putExtra(
android.content.Intent.EXTRA_TEXT, user);
targetedShareIntent
.setPackage("com.google.android.apps.plus");
startActivity(targetedShareIntent);
}
"v" is a button and social check if app is installed
any sugestion?

both options open google+ application?
I suppose show two options because google+ app have two activities with category.LAUNCHER.... the app and the chat app
PSDT: sorry for my bad english!

I was able to get the G+ dialog to launch with the following code (after querying the package manager to make sure the app was actually installed and passing in the currentContext as a parameter):
Intent appIntent = new Intent(Intent.ACTION_SEND);
String shareText = "Share text goes here";
appIntent.setType("text/plain");
appIntent.putExtra(Intent.EXTRA_TEXT, shareText);
//Filters so only the G+ app will launch
appIntent.setPackage("com.google.android.apps.plus");
try {
currentContext.startActivity(appIntent);
} catch (android.content.ActivityNotFoundException e) {
Log.d(e.getMessage());
}
Drove me bananas figuring this out, so I figured I'd try to save someone else the hassle.

Android: starting a default application or one which may be different on different devices

I would like to start a default application: browser, contact-book, phone, email, music app, etc. I have found many q/a, like browser opening a specific URL or blank, and here the answer is even "No not possible". But I would like to just open/launch it without telling it to go to a specific URL or sending a mail to someone, etc.
However, I also saw some Home applications where this seems to be working (at least for some apps). On my colleague's device there is for example a different contact-book (no google) which is detected and opened correctly.
I have seen in the Android documentation some intent categories that point to these problems, but these are only >= API.11. So I can't use/test them on my device.
Question: Is it not somehow possible to launch a default application (having the app chooser is of course ok) without providing extra data? If no, what do you think are these Home apps doing (perhaps workarounds are somehow possible).
PS: for the phone app I think, I have a workaround using Intent.ACTION_DIAL without any other information which will open simply the dialer.
UPDATE: I modified the title. Some applications like the address book may not be the same on different devices. So in this case I would like to start the address-book app, whichever this is.

This answer is not a 100% answer, but some workarounds on some typical applications.
Still open are: music player, address book
Browser: I get a list of applications that handle "http"-data intents, and then I look if one is available in the list of preferred applications.
Intent appFilter = new Intent(Intent.ACTION_VIEW);
appFilter.setData(Uri.parse("http://www.google.com"));
List<ResolveInfo> browserInfoList = pm.queryIntentActivities(appFilter, 0);
List<IntentFilter> outFilters = new ArrayList<IntentFilter>();
List<ComponentName> outActivities = new ArrayList<ComponentName>();
pm.getPreferredActivities(outFilters, outActivities, null);
if(outActivities.size() > 0) {
for(ComponentName cn : outActivities) {
String cnClass = cn.getClassName();
String cnPkg = cn.getPackageName();
for (ResolveInfo info : browserInfoList) {
if(info.activityInfo.name.equals(cnClass) &&
info.activityInfo.packageName.equals(cnPkg)) {
return cn;
}
}
}
}
In case no default is found, I open a browser chooser dialog, see here.
Phone: as described in the question:
Intent intent = new Intent(Intent.ACTION_DIAL);
startActivity(intent);

You can start apps by the function "startActivity" if you know about the canonical app name
like "android.com.browser". Do this simple by searching for AndroidManifest.xml in the app
source code (look at Codeaurora.com or at github/Cyanogenmod) and grab the app name you want.
After you know about the App name ("Activity") implement the code as follows:
Intent intent = new Intent();
intent.setClassName(this, "com.android.browser");
intent.setCategory(Intent.ACTION_MAIN);
startActivity(intent);
THIS is only a example, sometimes you have to put intent extras or data values, this information can be found in the app's AndroidManifest.xml too.