Is an Android Intent equivalent to a Java Event? - java

I'm trying to understand Android Intents and I understand the concept of a Java event and an EventListener. Are these similar or the the same?

Intent in Android, is used to express an "intention" as the name implies, to perform an action over another component. One common usage of intent is to use it along with startActivity(intentObject), where the object contains the information(contexts) of the starting and openning activity, see below:
Intent intent = new Intent(context, MyActivityB);//pass the information(contexts of the current activity and the activity I want to open).
startActivity(intent);//Use startActivity method to start the activity defined in the object intent.
Other actions include: openning an app to send an email, open a social media, etc.

Related

How to launch the google maps app from another app?

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

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:".

Get Chosen App from Intent.createChooser

I am trying to capture the result of Intent.createChooser to know which app a user selected for sharing.
I know there have been a lot of posts related to this:
How to know which application the user chose when using an intent chooser?
https://stackoverflow.com/questions/6137592/how-to-know-the-action-choosed-in-a-intent-createchooser?rq=1
How to get the user selection from startActivityForResult(Intent.createChooser(fileIntent, "Open file using..."), APP_PICKED);?
Capturing and intercepting ACTION_SEND intents on Android
but these posts are somewhat old, and I am hoping that there might be some new developments.
I am trying to implement a share action without having it be present in the menu. The closest solution to what I want is provided by ClickClickClack who suggest implementing a custom app chooser, but that seems heavy handed. Plus, it seems like there might be some Android hooks to get the chosen app, like the ActivityChooserModel.OnChooseActivityListener.
I have the following code in my MainActivity, but the onShareTargetSelected method is never getting called.
Intent sendIntent = new Intent();
sendIntent.setAction(Intent.ACTION_SEND);
sendIntent.putExtra(Intent.EXTRA_TEXT, shareMessage());
sendIntent.setType("text/plain");
Intent intent = Intent.createChooser(sendIntent, getResources().getText(R.string.share_prompt));
ShareActionProvider sap = new ShareActionProvider(this);
sap.setShareIntent(sendIntent);
sap.setOnShareTargetSelectedListener(new ShareActionProvider.OnShareTargetSelectedListener() {
#Override
public boolean onShareTargetSelected(ShareActionProvider source, Intent intent) {
System.out.println("Success!!");
return false;
}
});
startActivityForResult(intent, 1);
As of API level 22 it is now actually possible. In Android 5.1 a method (createChooser (Intent target, CharSequence title, IntentSender sender)) was added that allows for receiving the results of the user's choice. When you provide an IntentSender to createChooser, the sender will be notified by the chooser dialog with the ComponentName chosen by the user. It will be supplied in the extra named EXTRA_CHOSEN_COMPONENT int the IntentSender that is notified.
I am trying to capture the result of Intent.createChooser to know which app a user selected for sharing.
That is not possible.
Other "choosing" solutions, like ShareActionProvider, may offer more. I have not examined the Intent handed to onShareTargetSelected() to see if it contains the ComponentName of the chosen target, though the docs suggest that it should.
And, if for some reason it does not, you are welcome to try to fork ShareActionProvider to add the hooks you want.
The reason why createChooser() cannot be handled this way is simply because the "choosing" is being done by a separate process from yours.
I have the following code in my MainActivity, but the onShareTargetSelected method is never getting called.
ShareActionProvider goes in the action bar. You cannot just create an instance, call a couple of setters, and expect something to happen.

How to save an intent from one class to another class

I am new to Android so apologies if I am asking something silly. I am trying to develop an alarm clock application - basically, it's my final project and I am trying to develop an alarm like there is in API level 2.3.3.
I have designed the list view that takes input through a dialog box like time. I have also coded it to set an alarm.
Now I want to save that alarm as an intent in the other class, and I don't have any idea how to save different alarms in the other activity. I have also checked for the desk-clock alarm code but I didn't get that too.
Please help me someone, I am stuck here for the code for more than a week. Please someone help me, I shall be thankful to you.
If you want to send an Intent from one Activity to another, and then retrieve information from inside the Intent, the best way is use the Bundle object inside the intent:
Let's supose you send the intent from Activity1 to Activity2...
In Activity 1:
Intent intent = new Intent(Activity1.class,Activity2.class);
//I use the String class name as a key value, but you can use whatever key
intent.putExtra(String.class.getCanonicalName(), myString);
startActivity(intent);
//Or this other method if you want to retrieve a result from Activity2
//startActivityForResult(intent,Activity2);
In Activity 2:
Intent intent = this.getIntent();
Bundle bundle = intent.getExtras();
String myString = bundle.getString(String.class.getCanonicalName());

How do I create an android shortcut to an activity?

OK,I'm new at this forum, so don't blame me for putting this in the wrong tags,not putting something in,eg.
I want to learn how to create a shortcut(I did that by Googling) and link it to an activity (In this case, com.android.mms.ui.ComposeMessageActivity)
I tried doing it, but it only showed me a toast saying "Application not installed" and I'm pretty sure it is.
It would be better if you can display a "complete action with another application" dialog.
If I assumed your question correctly, you mean a button or something within an activity that leads to another activity, that being -- "com.android.mms.ui.ComposeMessageActivity"
if your activity that you want to link to is in another application-- then
Intent intent = new Intent();
intent.setComponent(new ComponentName("com.mine", "com.android.mms.ui.ComposeMessageActivity"));
startActivity(intent);
if it is within the same application, then
Intent intent = new Intent(this, ComposeMessageActivity.class);
startActivity(intent);
//optional add this to your manifest to finish the current loading activity so
//as to not keep it in the activity stack
//<activity android:name="yourActivity" android:noHistory="true" ... />
EDIT If you mean a shortcut on a homescreen, then I would create a tiny application that only has one activity which uses the above method to link to a different application. Then I would drag that application to the home screen, and boom. If there's a better way, then please feel free to correct me

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