I am developing a file uploading android application. My objective is to upload the user selected file from file manager to a remote server. But when a google drive file is selected , file uploading fails because of empty path . Can somebody help me ?
My code is :
private void showFileChooser() {
Intent intent = new Intent();
intent.setType("*/*");
intent.setAction(Intent.ACTION_GET_CONTENT);
intent.addCategory(Intent.CATEGORY_OPENABLE);
intent.putExtra(Intent.EXTRA_LOCAL_ONLY, true);
startActivityForResult(Intent.createChooser(intent, "Choose File to Upload.."), PICK_FILE_REQUEST);
}
#Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (resultCode == Activity.RESULT_OK) {
if (requestCode == PICK_FILE_REQUEST) {
if (data == null) {
//no data present
return;
}
Uri selectedFileUri = data.getData();
selectedFilePath = FilePath.getPath(mActivity, selectedFileUri);
if (selectedFilePath != null && !selectedFilePath.equals("")) {
callUploadDocumentAPI();
} else {
Toast.makeText(mActivity, StringConstants.CANT_UPLOAD_TO_SERVER, Toast.LENGTH_SHORT).show();
}
}
}
But when a google drive file is selected , file uploading fails because of empty path .
FilePath.getPath(mActivity, selectedFileUri) cannot work. A Uri is not a file.
Use a ContentResolver and openInputStream() to get an InputStream on the content identified by the Uri. Either use that InputStream directly, or use it to make your own copy of the content in some file that you control, then use that copy.
Related
I get pictures url from gallery or camera on android but I can't convert them to file. When i developing an app with Java in android, sometimes I need to get pictures from gallery and camera to my application. But when i try to do this I'm getting errors.
Isn't there a standard here? The code works at one api level and does not works another one. Or when I take the URI of some images and convert it to file, I get an error regardless of the api level. I wonder if there is a fundamental solution to this problem?
I am starting intent with this
Intent intent = new Intent();
intent.setType("image/*");
intent.setAction(Intent.ACTION_GET_CONTENT);
startActivityForResult(intent,IMAGE_RESULT);
And then I'm catching data with this code:
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (resultCode == Activity.RESULT_OK) {
if (requestCode == IMAGE_RESULT) {
String filePath = ImageFilePath.getPath(OwnProfileActivity.this, data.getData());
selectedFile = new File(filePath);
showProgressDialog();
Handler handler = new Handler();
handler.postDelayed(this::chaneProfileImage, 1000);
}
}
}
May I suggest something like this, but of course this is if you have a bitmap:
private fun getImageUriFromBitmap(image: Bitmap): Uri? {
val bytes = ByteArrayOutputStream()
image.compress(Bitmap.CompressFormat.PNG, 100, bytes)
val relativeLocation = Environment.DIRECTORY_PICTURES + File.pathSeparator + "NotifyMe"
val contentValues = ContentValues().apply {
put(MediaStore.MediaColumns.DISPLAY_NAME, "notifyMe_${System.currentTimeMillis()}")
put(MediaStore.MediaColumns.MIME_TYPE, "image/png")
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.Q) {
put(MediaStore.MediaColumns.RELATIVE_PATH, relativeLocation)
put(MediaStore.MediaColumns.IS_PENDING, 1)
}
}
return context?.contentResolver?.insert(MediaStore.Images.Media.EXTERNAL_CONTENT_URI, contentValues)
}
Then you can create the file with the uri
val file = File(uri.path)
source: https://gist.github.com/franciscobarrios/9f6634c87ca6e56ed70cd96807dd1dcb
I am trying to fetch a file this way:
final Intent chooseFileIntent = new Intent(Intent.ACTION_GET_CONTENT);
String[] mimetypes = {"application/pdf"};
chooseFileIntent.setType("*/*");
chooseFileIntent.addCategory(Intent.CATEGORY_OPENABLE);
if (chooseFileIntent.resolveActivity(activity
.getApplicationContext().getPackageManager()) != null) {
chooseFileIntent.putExtra(Intent.EXTRA_MIME_TYPES, mimetypes);
activity.startActivityForResult(chooseFileIntent, Uploader.PDF);
}
Then in onActivityResult :
#Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
}
According to many threads I'm supposed to fetch the file name from the intent with data.getData().getPath(), the file name I'm expecting is my_file.pdf, but instead I'm getting this :
/document/acc=1;doc=28
So what to do? Thanks for your help.
I am trying to fetch a file
Not with that code. That code is asking the user to pick a piece of content. This may or may not be a file.
According to many threads I'm supposed to fetch the file name from the intent with data.getData().getPath()
That was never correct, though it tended to work on older versions of Android.
So what to do?
Well, that depends.
If you wish to only accept files, integrate a file chooser library instead of using ACTION_GET_CONTENT. (UPDATE 2019-04-06: since Android Q is banning most filesystem access, this solution is no longer practical)
If you are willing to allow the user to pick a piece of content using ACTION_GET_CONTENT, please understand that it does not have to be a file and it does not have to have something that resembles a filename. The closest that you will get:
If getScheme() of the Uri returns file, your original algorithm will work
If getScheme() of the Uri returns content, use DocumentFile.fromSingleUri() to create a DocumentFile, then call getName() on that DocumentFile — this should return a "display name" which should be recognizable to the user
To get the real name and to avoid getting a name that looks like "image: 4431" or even just a number, you can write code as recommended by CommonsWare.
The following is an example of a code that selects a single pdf file, prints its name and path to the log, and then sends the file by email using its uri.
private static final int FILEPICKER_RESULT_CODE = 1;
private static final int SEND_EMAIL_RESULT_CODE = 2;
private Uri fileUri;
private void chooseFile() {
Intent fileChooser = new Intent(Intent.ACTION_GET_CONTENT);
fileChooser.setType("application/pdf");
startActivityForResult(Intent.createChooser(fileChooser, "Choose one pdf file"), FILEPICKER_RESULT_CODE);
}
#Override
protected void onActivityResult(int requestCode, int resultCode, #Nullable Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (requestCode == FILEPICKER_RESULT_CODE) {
if (resultCode == RESULT_OK) {
fileUri = data != null ? data.getData() : null;
if (fileUri != null) {
DocumentFile d = DocumentFile.fromSingleUri(this, fileUri);
if (d != null) {
Log.d("TAG", "file name: " + d.getName());
Log.d("TAG", "file path: " + d.getUri().getPath());
sendEmail(fileUri);
}
}
}
}
}
private void sendEmail(Uri path) {
String email = "example#gmail.com";
Intent intent = new Intent(android.content.Intent.ACTION_SEND);
intent.setType("application/octet-stream");
intent.putExtra(android.content.Intent.EXTRA_SUBJECT, "PDF file");
String[] to = { email };
intent.putExtra(Intent.EXTRA_EMAIL, to);
intent.putExtra(Intent.EXTRA_TEXT, "This is the pdf file...");
intent.putExtra(Intent.EXTRA_STREAM, path);
startActivityForResult(Intent.createChooser(intent, "Send mail..."), SEND_EMAIL_RESULT_CODE);
}
hope it helps.
In my application i want to create subdirectories in choosen directory. Im using SAF (Storage Access Framework): choose directory and create one subdirectory works fine.
Choose directory:
Intent intent = new Intent(Intent.ACTION_OPEN_DOCUMENT_TREE);
startActivityForResult(intent, REQUEST_CODE_PATH_TO_DATA);
Create subdirectory:
#Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (resultCode != Activity.RESULT_OK) {
return;
}
if (requestCode == REQUEST_CODE_PATH_TO_DATA) {
if (data == null) {
return;
}
DocumentFile pickedDir = DocumentFile.fromTreeUri(getActivity(), data.getData());
getActivity().grantUriPermission(getActivity().getPackageName(), data.getData(), Intent.FLAG_GRANT_READ_URI_PERMISSION | Intent.FLAG_GRANT_WRITE_URI_PERMISSION);
getActivity().getContentResolver().takePersistableUriPermission(data.getData(), Intent.FLAG_GRANT_READ_URI_PERMISSION | Intent.FLAG_GRANT_WRITE_URI_PERMISSION);
pickedDir.createDirectory("Portfolio");
String pathToPickedDir = "";
for (DocumentFile file : pickedDir.listFiles()) {
if (file.isDirectory() && file.getName().endsWith("Portfolio")) {
pathToPickedDir = file.getUri().toString();
}
}
}
}
URI from pickedDir is:
content://com.android.externalstorage.documents/tree/314E-7741%3ADCIM
And URI from created subdirectory "Portfolio" is:
content://com.android.externalstorage.documents/tree/314E-7741%3ADCIM/document/314E-7741%3ADCIM%2FPortfolio
And then when im trying to create subdirectory in "Portfolio" i can't do that, because the directory is created in the folder that was initially selected, not in the "Portfolio" folder.
DocumentFile pickedDir = DocumentFile.fromTreeUri(getActivity(), pathToPickedDir);
pickedDir.createDirectory("Patient");
What am I doing wrong? Thank you for your help.
Only way that i found is using "for" cycle:
for (DocumentFile file : pickedDir.listFiles()) {
if (file.isDirectory() && file.getName().equals("Portfolio")) {
file.createDirectory("Subdirectory");
}
}
The following should work in your case:
DocumentFile portfolioDir = pickedDir.createDirectory("Portfolio");
portfolioDir.createDirectory("Patient");
In your code, you are not using the directory object returned by DocumentFile.createDirectory(), which is the object of the newly created directory, in which you want to create a sub-directory.
I'm using file helper class from this post https://stackoverflow.com/a/20559175/2281821 but it doesn't working properly in each case. I'm using Intent chooseIntent = new Intent(Intent.ACTION_OPEN_DOCUMENT); to start file picker. I have Android Nougat and I'm receiving from onActivityResult uri: content://com.android.externalstorage.documents/document/home%3Aimage.png
What does this home: means? How can I get access to this file? I was trying with Environment.getExternalStorageDirectory() and System.getenv("EXTERNAL_STORAGE") but I can't get acess.
Following code can be used to get document path:
Intent galleryIntent = new Intent();
galleryIntent .setType("image/*");
galleryIntent .setAction(Intent.ACTION_GET_CONTENT);
startActivityForResult(Intent.createChooser(galleryIntent , getString(R.string.app_name)),REQUEST_CODE);
OnActivityResult():
#Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
if (resultCode == RESULT_OK) {
if (requestCode == REQUEST_CODE) {
Uri uri = data.getData();
}
}
}
You can use "uri" variable value to get bitmap. If you want to open document other image also then remove following line:
galleryIntent .setType("image/*");
You get access to a content scheme by opening for instance an InputStream on the uri.
InputStream is = getContentResolver().openInputStream(uri);
BitmapFactory.decodeFromStream(is) will happily read from your stream.
I'm making an app that picks a image from the gallery and returns the result. I need to get the file path from that selection and store it for later use. I have looked at similar questions and apparently the file path of the image is stored in the column
MediaStore.Images.Media.DATA
How would anyone know that this is the file path? I've looked at the android documentation on MediaStore and it does not tell you where the file path column is. Any help would be appreciated.
Intent i = new Intent(Intent.ACTION_PICK,android.provider.MediaStore.Images.Media.EXTERNAL_CONTENT_URI);
startActivityForResult(i, SELECT_PHOTO);
use this to get an image from gallery, it will return URI of the image which could be store store for later use
#Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (resultCode != RESULT_CANCELED) {
if (requestCode == SELECT_PHOTO) {
Uri path = data.getData();
}
}
}