Develop Reference

Convert android Uri to java File Android Studio

I have an Intent of ACTION_GET_CONTENT in my app and I need to put the picked file ( there will be different files, ppt, doc...) in a java.io File.
I'm able to get the data and put it into a android.net Uri. Is there a way I ca create a java File from this Uri?
I need it to be a file in order to upload it do google drive using the google drive API
This is the code to upload to the drive, I need to convert the uri to a temporary file in order to pass it as the javaFile of this method
public Task<File> uploadFileWithMetadata(java.io.File javaFile, boolean isSlide, #Nullable final String folderId, PostFileHolder postFileHolder) {
return Tasks.call(mExecutor, () -> {
Log.i("upload file", "chegou" );
String convertTo; // string to convert to gworkspace
if(isSlide){
convertTo = TYPE_GOOGLE_SLIDES;
}
else{
convertTo = TYPE_GOOGLE_DOCS;
}
List<String> folder;
if (folderId == null) {
folder = Collections.singletonList("root");
} else {
folder = Collections.singletonList(folderId);
}
File metadata = new File()
.setParents(Collections.singletonList(folderId))
.setName(postFileHolder.getDisplayName())
.setMimeType(convertTo);
Log.i("convert to: ", convertTo );
// the convert to is the mimeType of the file, withg gworkspace it is a gdoc or gslide, with others is the regular mimetype
FileContent mediaContent = new FileContent(postFileHolder.getConvertTo(), javaFile);
Log.i("media content", "chegou" );
// até aqui com gworkspace chega
File uploadedFile = mDriveService.files().create(metadata, mediaContent)
.setFields("id")
.execute();
Log.i("File ID: " , uploadedFile.getId());
return uploadedFile;
});
}
This is my code to get the Uri
case REQUEST_CODE_FILE_PICKER:
// get uri from file picked
Uri url = data.getData();
break;
}

Solved it!
Here's how I did it:
// my uri
Uri fileUri = Uri.parse(postFileHolder.getFileUri());
// create a null InputSream
InputStream iStream = null;
try {
// create a temporary file
File fileToUpload = File.createTempFile("fileToUpload", null, this.getCacheDir());
iStream = getContentResolver().openInputStream(fileUri);
// use function to get the bytes from the created InputStream
byte[] byteData = getBytes(iStream);
convert byteArray to File
FileOutputStream fos = new FileOutputStream(fileToUpload);
fos.write(byteData);
fos.flush();
fos.close();
if(fileToUpload == null){
Log.i("create file", "null");
}
else{
Log.i("create file", "not null: "+ fileToUpload.getTotalSpace());
getEGDrive(fileToUpload);
}
}
catch (FileNotFoundException e) {
Log.i("error create file uri", e.getLocalizedMessage());
e.printStackTrace();
} catch (IOException e) {
Log.i("error create file uri", e.getLocalizedMessage());
e.printStackTrace();
}
And here's the function to transform the InputStream into byteArray:
public byte[] getBytes(InputStream inputStream) throws IOException {
ByteArrayOutputStream byteBuffer = new ByteArrayOutputStream();
int bufferSize = 1024;
byte[] buffer = new byte[bufferSize];
int len = 0;
while ((len = inputStream.read(buffer)) != -1) {
byteBuffer.write(buffer, 0, len);
}
return byteBuffer.toByteArray();
}
Got big part of the answer from: https://stackoverflow.com/a/10297073/14990708

Reading a File line by line

I'm working on the creation of an Android App to track financial expenses and I would like to allow the user to batch import financial transactions into my app. The transactions would be come as .csv/.txt files.
However, I'm getting a cryptic exception:
java.lang.AbstractMethodError: abstract method "android.content.IContentProvider android.content.ContentResolver.acquireUnstableProvider(android.content.Context, java.lang.String)"
at android.content.ContentResolver.acquireUnstableProvider(ContentResolver.java:1780)
at android.content.ContentResolver.openTypedAssetFileDescriptor(ContentResolver.java:1394)
at android.content.ContentResolver.openAssetFileDescriptor(ContentResolver.java:1247)
at android.content.ContentResolver.openInputStream(ContentResolver.java:967)
....
Workflow is as follow: User selects text file to import, App imports content.
Start the file selector:
Intent intent = new Intent(Intent.ACTION_OPEN_DOCUMENT);
intent.setType("*/*");
//intent.addCategory("CATEGORY_OPENABLE");
startActivityForResult(intent, REEQUEST_CODE_IMPORT);
Catch the result:
public void onActivityResult(int requestCode, int resultCode, Intent data) {
// Check which request we're responding to
if (requestCode == REEQUEST_CODE_IMPORT) {
// Make sure the request was successful
Uri path = data.getData();
if (resultCode == RESULT_OK && path != null) {
InputStream inputStream = null;
try {
ContentResolver contentResolver = new ContentResolver(getContext()) {};
// Error happens in the next line
inputStream = contentResolver.openInputStream(path);
// Get the object of DataInputStream
DataInputStream in = new DataInputStream(inputStream);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
String line = "";
while((line = br.readLine()) != null) {
// Do something meaningful...
}
} catch () {
// Catch the exceptions ( I have removed some boiler plate code here...)
} finally {
// Close the path ( I have removed some boiler plate code here...)
inputStream.close();
}
}
}
}

Android documentation helped me solve the issue: Access documents and other files from shared storage - Input Stream
Following code adaptation is working:
public void onActivityResult(int requestCode, int resultCode, Intent data) {
// Check which request we're responding to
if (requestCode == REEQUEST_CODE_IMPORT) {
// Make sure the request was successful
Uri path = data.getData();
if (resultCode == RESULT_OK && path != null) {
StringBuilder stringBuilder = new StringBuilder();
try (InputStream inputStream = getActivity().getContentResolver().openInputStream(path);
BufferedReader reader = new BufferedReader(
new InputStreamReader(Objects.requireNonNull(inputStream)))) {
String line;
while ((line = reader.readLine()) != null) {
Log.d(TAG, line);
stringBuilder.append(line);
}
} catch (IOException e) {
e.printStackTrace();
}
String content = stringBuilder.toString();
// Do something with it
}
}
}

Using SAF File Picker in Android 10 (Q) to copy files from Downloads to local app folder

Since Android 10 (API 29) I need to use the Storage Access Framework's File Picker to select GPX (GPS) files to copy from the Downloads folder to my local app folder. I have implemented the file picker and am able to select the GPX file, however the result data URI appears different to the filename (but unique) and I cannot seem to use it to copy the files. The rest of the code is the same "copy" code I used in previous versions of Android. What am I doing wrong and how should I best use the SAF File Picker to copy files? I haven't been able to find a recent (API 29) "file copy" example on the net...
private static final int READ_REQUEST_CODE = 42;
...
public void performFileSearch() {
Intent intent = new Intent(Intent.ACTION_OPEN_DOCUMENT);
intent.addCategory(Intent.CATEGORY_OPENABLE);
// intent.setType("application/gpx"); // Filters GPX file but wont let me select them.
intent.setType("*/*");
startActivityForResult(intent, READ_REQUEST_CODE);
}
...
if (requestCode == READ_REQUEST_CODE && resultCode == Activity.RESULT_OK) {
Uri uri = null;
if (data != null) {
uri = data.getData();
handleDownloadedGPXFiles2(uri);
}
}
...
private void handleDownloadedGPXFiles2(Uri selectedFileUri) {
File sourceFile = new File(selectedFileUri.getPath()); // Returns a unique number or string but NOT filename string???
File destDirectory = new File(this.getExternalFilesDir(null), "Imported");
File destFile = new File(destDirectory, "test.gpx"); // Needs to be same name as original filename.
try {
if (!destFile.exists()) {
destFile.createNewFile();
}
FileInputStream inStream = new FileInputStream(sourceFile);
FileOutputStream outStream = new FileOutputStream(destFile);
FileChannel inChannel = inStream.getChannel();
FileChannel outChannel = outStream.getChannel();
inChannel.transferTo(0, inChannel.size(), outChannel);
inStream.close();
outStream.close();
} catch (IOException e) {
e.printStackTrace();
}
Toast.makeText(getApplicationContext(), "File Import Complete", Toast.LENGTH_LONG).show();
}

File sourceFile = new File(selectedFileUri.getPath());
Remove above line.
FileInputStream inStream = new FileInputStream(sourceFile);
Replace that line by:
InputStream inStream = getContentResolver().openInputStream(selectedFileUri);
Further you can remove
if (!destFile.exists()) {
destFile.createNewFile();
}
as the file will be created by the new FileOutputStream();
Finally: Your last Toast() is on the wrong place. It should be in the try block.
Place a different Toast() in the catch block to inform yourself or the user.

Thanks blackapps. Final code works well...
private void handleDownloadedGPXFiles2(Uri selectedFileUri) {
String displayName = "imported.gpx";
String fileExtension;
ContentResolver contentResolver = getContentResolver();
Cursor cursor = contentResolver.query(selectedFileUri, null, null, null, null);
try {
if (cursor != null && cursor.moveToFirst()) {
displayName = cursor.getString(cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
if (displayName != null && displayName.length() >=4) {
fileExtension = displayName.substring(displayName.length() - 4);
if (!fileExtension.equals(".gpx")){
myCustomToast("Must be a .GPX file!");
return;
}
} else {
myCustomToast("Must be a .GPX file!");
return;
}
}
File destDirectory = new File(this.getExternalFilesDir(null), "Imported");
File destFile = new File(destDirectory, displayName);
FileOutputStream outStream = new FileOutputStream(destFile);
InputStream in = getContentResolver().openInputStream(selectedFileUri);
OutputStream out = outStream;
byte[] buffer = new byte[1024];
int read;
while ((read = in.read(buffer)) != -1) {
out.write(buffer, 0, read);
}
in.close();
out.flush();
out.close();
Toast.makeText(getApplicationContext(), "File Import Complete", Toast.LENGTH_LONG).show();
} catch (IOException e) {
Toast.makeText(getApplicationContext(), "File Import FAILED", Toast.LENGTH_LONG).show();
e.printStackTrace();
}
finally
{
if (cursor != null)
cursor.close();
}
}

Access a text file in android

I'm trying to read the contents of a flat file stored in the phone's memory. But when I get the file address through the attempt, and I create the File object this is not readable.
Try to create it by manually passing the file's address and it works.
The problem is that I can not get that address from the uri that gives me the Intent
Here I open the android file browser
btnAdjuntar.setOnClickListener(new View.OnClickListener() {
#Override
public void onClick(View v) {
Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
intent.setType("text/plain");
Intent chooser = Intent.createChooser(intent, "Elige App");
if (chooser.resolveActivity(getPackageManager()) != null) {
startActivityForResult(chooser, COD_ABRIR_FICHERO);
}
}
});
Here I receive the Intent
#Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (requestCode == COD_ABRIR_FICHERO) {
if (resultCode == RESULT_OK) {
String ruta = data.getData().getPath();
String lectura;
try {
lectura = leerArchivo(ruta);
} catch (Exception e) {
lectura = e.getMessage();
}
Toast.makeText(getApplicationContext(), lectura, Toast.LENGTH_LONG).show();
}
}
}
The exception is thrown and you have this message:
/document/primary:Documents/PruebaArchivo.txt: open failed: ENOENT (No such file or directory)
Method leerArchivo
public String leerArchivo(String ruta) throws FileNotFoundException, IOException {
String cadena;
String lectura = "";
File file = new File(ruta);
FileReader f = new FileReader(file);
BufferedReader b = new BufferedReader(f);
while ((cadena = b.readLine()) != null) {
lectura += cadena;
}
b.close();
return lectura;
}
Already the permissions to read the external memory are accepted

You can open an InputStream from a Uri:
InputStream is = getContentResolver().openInputStream(uri);
Then wrap the input stream to create a BufferedReader:
BufferedReader b = new BufferedReader(new InputStreamReader(is));
while ((cadena = b.readLine()) != null) {
lectura += cadena;
}
b.close();
etc.
This way you never need to get the file's address. You can specify the encoding (e.g. "UTF-8") as a second parameter to the InputStreamReader constructor.

But when I get the file address through the attempt, and I create the File object this is not readable.
You are not getting a file. You are getting content. You are being given a Uri. A Uri is not a file.
String ruta = data.getData().getPath();
That line is only useful if the scheme of the Uri is file. In your case, the scheme is probably content.
Use a ContentResolver and openInputStream() to get an InputStream on the content. This works whether the Uri has a file scheme or a content scheme.

How to get Base64 String from InputStream?

I'm on a problem by taking the selected gallery picture and want to save it first as Base64 String in a XML file (for later use. For example if you exit the app and open it again).
As you can see I get the Image on a InputStream
But first of all the onClick method:
public void onClick(DialogInterface dialog, int which) {
pictureActionIntent = new Intent(Intent.ACTION_GET_CONTENT);
pictureActionIntent.setType("image/*");
startActivityForResult(pictureActionIntent,GALLERY_PICTURE);
}
Now in the onActivityResult method I want to store the image from InputStream to Base64 String.
case GALLERY_PICTURE:
if (resultCode == RESULT_OK && null != data) {
InputStream inputstream = null;
try {
inputstream = getApplicationContext().getContentResolver().openInputStream(data.getData());
Base64InputStream in = new Base64InputStream(inputstream,0);
} catch (IOException e) {
e.printStackTrace();
}
#EDIT
This is what I do after creating the base64 String.
Bitmap bmp = base64EncodeDecode.decodeBase64(Items.get("image"));
Image1.setImageBitmap(bmp);
And this is the decoding Method:
public Bitmap decodeBase64(String input) {
byte[] decodedByte = Base64.decode(input, Base64.DEFAULT);
return BitmapFactory.decodeByteArray(decodedByte, 0, decodedByte.length);
}
I tried to use Base64InputStream but without success.
Can you give me a hint how to get from InputStream to Base64 String?
How many steps it will take doesn't matter.
I hope someone can help me!
Kind Regards!

Write these lines in onActivityResult method
try {
// get uri from Intent
Uri uri = data.getData();
// get bitmap from uri
Bitmap bitmap = MediaStore.Images.Media.getBitmap(getContentResolver(), uri);
// store bitmap to file
File filename = new File(Environment.getExternalStorageDirectory(), "imageName.jpg");
FileOutputStream out = new FileOutputStream(filename);
bitmap.compress(Bitmap.CompressFormat.JPEG, 60, out);
out.flush();
out.close();
// get base64 string from file
String base64 = getStringImage(filename);
// use base64 for your next step.
} catch (IOException e) {
e.printStackTrace();
}
private String getStringImage(File file){
try {
FileInputStream fin = new FileInputStream(file);
byte[] imageBytes = new byte[(int)file.length()];
fin.read(imageBytes, 0, imageBytes.length);
fin.close();
return Base64.encodeToString(imageBytes, Base64.DEFAULT);
} catch (Exception ex) {
Log.e(tag, Log.getStackTraceString(ex));
toast("Image Size is Too High to upload.");
}
return null;
}
you can use base64 String of image.
Also don't forget to add permissions in AndroidManifest.xml file READ_EXTERNAL_STORAGE and WRITE_EXTERNAL_STORAGE
EDIT:: Decode base64 to bitmap
byte[] bytes = Base64.decode(base64.getBytes(), Base64.DEFAULT);
ImageView image = (ImageView) this.findViewById(R.id.ImageView);
image.setImageBitmap(
BitmapFactory.decodeByteArray(bytes, 0, bytes.length)
);
Hope it'll work.

This should work:
public static byte[] getBytes(Bitmap bitmap) {
try{
ByteArrayOutputStream stream = new ByteArrayOutputStream();
stream.flush();
//bitmap.compress(CompressFormat.PNG, 98, stream);
bitmap.compress(CompressFormat.JPEG, 98, stream);
//bitmap.compress(Bitmap.CompressFormat.JPEG, 90, stream);
return stream.toByteArray();
} catch (Exception e){
return new byte[0];
}
}
public static String getString(Bitmap bitmap){
byte [] ba = getBytes(bitmap);
String ba1= android.util.Base64.encodeToString(ba, android.util.Base64.DEFAULT);
return ba1;
}
Got this code from something I use in an application, stripped it down to the most basic as far as i know.

If you are selecting image from Gallery then why you are saving it as Base64 string in xml file , you can reuse that image from gallery .
For this save image url in SharedPreferences and use that url again to show image .
Edit :
If you want to store it locally then you can use SQLite Database to store it , for more detail visit this link .