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:".
Related
I've added a button to my application which is supposed to open the download folder of the phone, and from there you should be able to click on files that were stored there, from the same app. Right now im saving some data there.
Problem is; I cant open the saved files in the folder.
I can see the files stored right there, but when I press one of them you immediatley go back to the app and not the file that you pressed.
Is there something I'm missing? Are you not supposed to open files stored in external storage from another app?
I've tried adding permissions in manifest and checkSelfpermission for checks in runtime, but with no success.
Here's the button for opening download folder:
private void openSavedLocation(){
if (ContextCompat.checkSelfPermission(ExportAndImport.this, Manifest.permission.READ_EXTERNAL_STORAGE) != PackageManager.PERMISSION_GRANTED)
{
ActivityCompat.requestPermissions(ExportAndImport.this, new String[] {Manifest.permission.READ_EXTERNAL_STORAGE}, 1);
}
Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
Uri uri = Uri.parse(Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS).getPath());
intent.setDataAndType(uri, "text/xml");
startActivity(Intent.createChooser(intent, "Open Folder"));}
I can open the file perfectly when Im opening it outside the app, not via this "createChooser". What could i be missing?
Any help is appreciated.
but when I press one of them you immediatley go back to the app and not the file that you pressed
That is what your code does. ACTION_GET_CONTENT says "let the user pick a piece of content". It does not say "open that piece of content in some other app". There is no single Intent action for saying "let the user pick a piece of content, then open that piece of content in some other app".
Is there something I'm missing?
If you want to try to open the XML in some other app:
Use startActivityForResult(), not startActivity(), for your ACTION_GET_CONTENT request (and get rid of the createChooser() bit)
Override onActivityResult() to get the result of the user's choice
If the user chose something (i.e., you get RESULT_OK in onActivityResult()), create an ACTION_VIEW Intent wrapped around the Uri that you get from the Intent passed into onActivityResult(), and call startActivity() on the ACTION_VIEW Intent
If, instead, your objective is to open this XML in your app, you would:
Use startActivityForResult(), not startActivity(), for your ACTION_GET_CONTENT request (and get rid of the createChooser() bit)
Override onActivityResult() to get the result of the user's choice
If the user chose something (i.e., you get RESULT_OK in onActivityResult()), get the Uri of the content from the Intent passed into onActivityResult(), then use ContentResolver to do something useful with that Uri (e.g., openInputStream() to read in the content)
Here's the button for opening download folder
ACTION_GET_CONTENT uses the MIME type. It will not necessarily honor your supplied starting Uri.
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.
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);
}
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 you make it so if a user does something, it asks them how they want to do it and it brings them to that application? Like if you go "Share" a photo or video it pops up a list with all apps that can do that for you like facebook or text or email , ect.? I am asking how to have my app use other apps, not them use mine.
You do this by targeting the intent filter that the other apps use.
For example, if I want to share a picture, I can do this:
Intent share = new Intent(Intent.ACTION_SEND);
share.setType("image/jpeg");
share.putExtra(Intent.EXTRA_STREAM, Uri.parse("/sdcard/test.jpg"));
startActivity(Intent.createChooser(share, "Share image"));
Then you will get a popup to choose the app to use for this action.