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Android open external IMDB application from my app

I have a button in my application which opens the imdb application in the phone with a imdb id I received from https://developers.themoviedb.org/3/getting-started/introduction
But I couldnt find anyway(using intents) to make my app recognize the imdb app and open it and if imdb app do not exist then I want to open the web site. How can I accomplish this?

I think I may be able to point you in the right direction. Just to be sure, you seem to be using TMDB but wish to open in the IMDB app?
The code below is from the Android documentation.
It will start your intent if the package manager can find an app with the appropriate intent filter installed on your device. If multiple apps are able to open this intent then an app chooser should pop up, unless the user has previously set a default for this kind of URI.
Intent sendIntent = new Intent(Intent.ACTION_SEND);
// Always use string resources for UI text.
// This says something like "Share this photo with"
String title = getResources().getString(R.string.chooser_title);
// Create intent to show the chooser dialog
Intent chooser = Intent.createChooser(sendIntent, title);
// Verify the original intent will resolve to at least one activity
if (sendIntent.resolveActivity(getPackageManager()) != null) {
startActivity(chooser);
}
If you add an else onto that then you can use a view intent like this :
Intent internetIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(movieUrl));
//Watch out here , There is a URI and Uri class!!!!
if (internetIntent.resolveActivity(getPackageManager()) != null){
startActivity(internetIntent);
}
I also found this (rather old) post about calling an explicit imdb uri
Imdb description
startActivity(android.intent.action.VIEW, imdb:///title/<titleID>);
// take note of the Uri scheme "imdb"
I hope this helps. If you post some more detail , code, links , I might be able to work through this with you.
If my answer is way off base then please be kind and set me right. We are all learning every day!
Good Luck.

Android direct shared

I am trying to share a link from my app with direct share. The share dialog must be like the image below with the most used contacts from messaging apps, like WhatsApp contacts.
This is the Intent structure which I am using for share the link:
Intent shareIntent = ShareCompat.IntentBuilder
.from(getActivity())
.setType("text/plain")
.setText(sTitle+ "\n" + urlPost)
.getIntent();
if (shareIntent.resolveActivity(
getActivity().getPackageManager()) != null)
startActivity(shareIntent);
And this is what my app shows:
Any idea how to achieve that?

You should use .createChooserIntent() instead of .getIntent()

Like this code below, you can use Intent.createChooser
Intent sharingIntent = new Intent(Intent.ACTION_SEND);
Uri screenshotUri = Uri.parse("file://" + filePath);
sharingIntent.setType("image/png");
sharingIntent.putExtra(Intent.EXTRA_STREAM, screenshotUri);
startActivity(Intent.createChooser(sharingIntent, "Share image using"));

You should use .createChooserIntent() instead of .getIntent()
Docs: This uses the ACTION_CHOOSER intent, which shows
an activity chooser, allowing the user to pick what they want to before proceeding. This can be used as an alternative to the standard activity picker that is displayed by the system when you try to start an activity with multiple possible matches, with these differences in behavior:
You can specify the title that will appear in the activity chooser.
The user does not have the option to make one of the matching activities a preferred activity, and all possible activities will
always be shown even if one of them is currently marked as the
preferred activity.
This action should be used when the user will naturally expect to
select an activity in order to proceed. An example if when not to use
it is when the user clicks on a "mailto:" link. They would naturally
expect to go directly to their mail app, so startActivity() should be
called directly: it will either launch the current preferred app, or
put up a dialog allowing the user to pick an app to use and optionally
marking that as preferred.
In contrast, if the user is selecting a menu item to send a picture
they are viewing to someone else, there are many different things they
may want to do at this point: send it through e-mail, upload it to a
web service, etc. In this case the CHOOSER action should be used, to
always present to the user a list of the things they can do, with a
nice title given by the caller such as "Send this photo with:".

Initiliazing ParseLoginUI?

Where does one actually place the code to launch the ParseLoginUI activity?
ParseLoginBuilder builder = new ParseLoginBuilder(MainActivity.this);
startActivityForResult(builder.build(), 0);
Is it in the ParseLoginDispatchActivity? This was not made very clear at all within any of the official documentation:
https://github.com/ParsePlatform/ParseUI-Android
https://www.parse.com/docs/android/guide#user-interface
I'm importing ParseLoginUI into my existing app. What do I once I've installed everything, updated my manifests, my build.gradle and now want to actually launch the Login activity once my app launches?
Do I put something in my manifest to indicate that the ParseLoginActivity should launch first? That doesn't seem to work as an Activity from my main application is required to launch as the initial intent. I'm a little lost here... Any thoughts?

Well I did find one solution, albeit a trivial one:
Intent loginIntent = new Intent(MainActivity.this, ParseLoginActivity.class); startActivity(loginIntent);
I launched the above Intent with an options menu item, but you could do it with a button or whatever else suits your needs.
If you're importing ParseLoginUI into an existing app, it appears you can just launch ParseLoginActivity with a simple Intent. I wish they mentioned this on their integration tutorial. Seems like the most straightforward way to get it running.
This solution definitely launches the Activity you want, but it doesn't check for whether the user is logged in or not and hence doesn't redirect you to the appropriate pages in your log-in flow (which I believe has more to do with your Manifest). It does, however, allow you to successfully register a user and log in with Parse, which is a great start.

A better solution would be to add the following to the onCreate method in the Activity that launches when your app launches. So if when your app launches you land on FirstActivity, the following will check to see if you are logged in. If you are not, you will be sent the login screen, and if you are logged in you will be sent to the second Activity, which is presumably where your users will want to be when they open your app.
ParseUser currentUser = ParseUser.getCurrentUser();
if (currentUser != null) {
Intent launchMainActivity = new Intent(this, SecondActivity.class);
startActivity(launchMainActivity );
} else {
ParseLoginBuilder builder = new ParseLoginBuilder(FirstActivity.this);
startActivityForResult(builder.build(), 0);
}

Opening apps on buttons

Im trying to open. Google Search on my application. But the problem is. when I click the button. the COMPLETE ACTION USING windows is popping up instead of the google search.. search the net for more than an hour but it seems I cant find the solution.. here is my code..
#Override
public void onClick(View view) {
Intent intent = new Intent("android.intent.action.MAIN");
intent.setComponent(ComponentName.unflattenFromString("com.google.android.googlequicksearchbox"));
intent.addCategory("android.intent.category.LAUNCHER");
startActivity(intent);

This is due to Android's nature to allow users to make their own decisions. If, for instance, they have Bing installed and prefer it as a search engine over Google, they will have the option to select it. As far as I know, there is no way to open a specific app this way. If the user selects Google and makes it the default app for this action, it will auto-open in successive times, but they must make that conscious decision first.

have you probably tried to follow this guide..?
It describes well about what you might need..
hope it helps..

List<ResolveInfo> resolveInfoList = getPackageManager().queryIntentActivities(intent, 0);
for(int i = 0; i < resolveInfoList.size(); i++) {
ResolveInfo info = resolveInfoList.get(i);
if (info.activityInfo.packageName.equals("com.google.android.googlequicksearchbox") {
// that's the one you want
}
}
I don't have Eclipse available right now to test it, but this should help you get there. Also check out the documentation for queryIntentActivities()
PS: I'm not sure about that packageName for google search

Try this. This code will search the meaning of value of query variable on google.
String url = "http://www.google.com/search?q=" + "Meaning of " +query;
Intent intentSearch = new Intent(Intent.ACTION_VIEW);
intentSearch.setData(Uri.parse(url));
startActivity(intentSearch);

Android disable display off while charging

In my application I need to disable display power off when device is charging. There is an option in Developer Menu to disable it, so I can to send Intent for user to enable it.
Also I've found info about PowerManager and WakeLocks, but it is for Alarms, I think. And I must to handle, is device charging.
What is the better, or is there another way to do this?

I've do this by that code:
final boolean isStayAwake = isStayAwakeEnabled(context);
if (!isStayAwake) {
intent = new Intent(ACTION_APPLICATION_DEVELOPMENT_SETTINGS);
}
context.startActivity(intent);
There user must check 'stay awake' option
I've used my own ACTION_APPLICATION_DEVELOPMENT_SETTINGS constant because of problems with default one, which have not "com." prefix

Perhaps you want to try FLAG_KEEP_SCREEN_ON from the WindowManager.LayoutParams