Create and Share a File (Also a mysterious error)
- by Kirk
My goal is to create a XML file 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.
Chooser Activity Leak - Android
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;
}
