I'm trying to get every launchable applications in my device using this method:
apps = new ArrayList<>();
Intent intent = new Intent(Intent.ACTION_MAIN, null);
intent.addCategory(Intent.CATEGORY_LAUNCHER);
List<ResolveInfo> availableActivities = manager.queryIntentActivities(intent, 0);
for(ResolveInfo ri:availableActivities){
AppDetail app = new AppDetail();
app.label = ri.loadLabel(manager);
app.name = ri.activityInfo.packageName;
app.icon = ri.activityInfo.loadIcon(manager);
apps.add(app);
}
I tried to print the label and package name of those application and found this:
Contacts com.sonyericsson.android.socialphonebook
Phone com.sonyericsson.android.socialphonebook
They have different app label yet the same package name. When I tried to open the apps, both of them open Contact app.
Is there any way to differentiate them? Or did I use a wrong method to get list of application?
queryIntentActivities retrieves all activities that can be performed for the given intent. So it can returns activites info with same package name.
I believe that Phone and Contacts are the same Contact app.
Two different icons can be created for the same program, one for each different activity. This makes sense, since the MAIN/LAUNCHER intent filter essentially tells android that the activity is the app's starting activity. So if you add this filter to two activities it will give you two icons for the same app to enter different activities. Nothing in android's intent filter model forces each app to have one and only one starting activity.
Related
How can one launch google maps app from another app with no action specified , like there are answers but all of one them specify the action to do, but if we just want to launch and let the user decide what he wants to do, how can we implement that ?
By creating an intent like this
// Create a Uri from an intent string. Use the result to create an Intent.
val gmmIntentUri = Uri.parse("google.streetview:cbll=46.414382,10.013988")
// Create an Intent from gmmIntentUri. Set the action to ACTION_VIEW
val mapIntent = Intent(Intent.ACTION_VIEW, gmmIntentUri)
// Make the Intent explicit by setting the Google Maps package
mapIntent.setPackage("com.google.android.apps.maps")
// Attempt to start an activity that can handle the Intent
startActivity(mapIntent)
You can check google maps intents through the official documentation here https://developers.google.com/maps/documentation/urls/android-intents#kotlin
In my homework I need to get all the activities that are registered to the Android operating system in a list. After that I need to filter them by having category attribute "android.intent.category.LAUNCHER" and action "android.intent.action.MAIN".
I read this question (How to list all activities exposed by an application?) and about Package Manager, but all the methods in the Package Manager only return list of the activities registered to the application package.
As I couldn't find any resource or method saying that I can get all the activities, I'm getting suspicious that I can't do that.
Is there any way to perform this kind of action?
The following will give you list of packages category attribute android.intent.category.LAUNCHER and action android.intent.action.MAIN.
Intent mainIntent = new Intent(Intent.ACTION_MAIN, null);
mainIntent.addCategory(Intent.CATEGORY_LAUNCHER);
List<ResolveInfo> pkgAppsList = getPackageManager().queryIntentActivities(mainIntent, 0);
for (int i = 0; i < pkgAppsList.size(); i++) {
Log.e("Activity package", pkgAppsList.get(i).activityInfo.name);
}
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.
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:".
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);
}