My goal is to create a XML file on internal storage and then send it through the share Intent.
I’m able to create a XML file using this code
FileOutputStream outputStream = context.openFileOutput(fileName, Context.MODE_WORLD_READABLE);
PrintStream printStream = new PrintStream(outputStream);
String xml = this.writeXml(); // get XML here
printStream.println(xml);
printStream.close();
I’m stuck trying to retrieve a Uri to the output file in order to share it. I first tried to access the file by converting the file to a Uri
File outFile = context.getFileStreamPath(fileName);
return Uri.fromFile(outFile);
This returns file:///data/data/com.my.package/files/myfile.xml but I cannot appear to attach this to an email, upload, etc.
If I manually check the file length, it’s proper and shows there is a reasonable file size.
Next I created a content provider and tried to reference the file and it isn’t a valid handle to the file. The ContentProvider doesn’t ever seem to be called a any point.
Uri uri = Uri.parse("content://" + CachedFileProvider.AUTHORITY + "/" + fileName);
return uri;
This returns content://com.my.package.provider/myfile.xml but I check the file and it’s zero length.
How do I access files properly? Do I need to create the file with the content provider? If so, how?
Update
Here is the code I’m using to share. If I select Gmail, it does show as an attachment but when I send it gives an error Couldn’t show attachment and the email that arrives has no attachment.
public void onClick(View view) {
Log.d(TAG, "onClick " + view.getId());
switch (view.getId()) {
case R.id.share_cancel:
setResult(RESULT_CANCELED, getIntent());
finish();
break;
case R.id.share_share:
MyXml xml = new MyXml();
Uri uri;
try {
uri = xml.writeXmlToFile(getApplicationContext(), "myfile.xml");
//uri is "file:///data/data/com.my.package/files/myfile.xml"
Log.d(TAG, "Share URI: " + uri.toString() + " path: " + uri.getPath());
File f = new File(uri.getPath());
Log.d(TAG, "File length: " + f.length());
// shows a valid file size
Intent shareIntent = new Intent();
shareIntent.setAction(Intent.ACTION_SEND);
shareIntent.putExtra(Intent.EXTRA_STREAM, uri);
shareIntent.setType("text/plain");
startActivity(Intent.createChooser(shareIntent, "Share"));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
break;
}
}
I noticed that there is an Exception thrown here from inside createChooser(…), but I can’t figure out why it’s thrown.
E/ActivityThread(572): Activity
com.android.internal.app.ChooserActivity has leaked IntentReceiver
com.android.internal.app.ResolverActivity$1@4148d658 that was
originally registered here. Are you missing a call to
unregisterReceiver()?
I’ve researched this error and can’t find anything obvious. Both of these links suggest that I need to unregister a receiver.
- ChooserActivity has leaked IntentReceiver
- Why does Intent.createChooser() need a BroadcastReceiver and how to implement?
I have a receiver setup, but it’s for an AlarmManager that is set elsewhere and doesn’t require the app to register / unregister.
Code for openFile(…)
In case it’s needed, here is the content provider I’ve created.
public ParcelFileDescriptor openFile(Uri uri, String mode) throws FileNotFoundException {
String fileLocation = getContext().getCacheDir() + "/" + uri.getLastPathSegment();
ParcelFileDescriptor pfd = ParcelFileDescriptor.open(new File(fileLocation), ParcelFileDescriptor.MODE_READ_ONLY);
return pfd;
}
It is possible to expose a file stored in your apps private directory via a ContentProvider. Here is some example code I made showing how to create a content provider that can do this.
Manifest
In your ContentProvider override openFile to return the ParcelFileDescriptor
Make sure you have copied your xml file to the cache directory
Now any other apps should be able to get an InputStream for your private file by using the content uri (content://com.example.prov/myfile.xml)
For a simple test, call the content provider from a seperate app similar to the following