In android, when you get the uri from the gallery, it's value will be start with content://blahblahblah.blahblah.format, but if you get the uri from your phone's camera, it will be starting with file:///
Below is what I want to do:
private File uriToBitmap(Uri uri, int maxSize) throws FileNotFoundException {
try {
imageStream = getContentResolver().openInputStream(uri);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Bitmap claimBitmap = BitmapFactory.decodeStream(imageStream);
}
In this method I would like to pass a file type uri and use getContentResolver() function, but unfortunately the claimBitmap is a null, does that means the getContentResolver() method doens't accept the file type uri? Please help.
No, file scheme is supported by ContentResolver, if the uri which scheme is not support, this method will throw a FileNotFoundException. Please check the
public final InputStream openInputStream(Uri uri)
throws FileNotFoundException {
String scheme = uri.getScheme();
if (SCHEME_ANDROID_RESOURCE.equals(scheme)) {
// Note: left here to avoid breaking compatibility. May be removed
// with sufficient testing.
OpenResourceIdResult r = getResourceId(uri);
try {
InputStream stream = r.r.openRawResource(r.id);
return stream;
} catch (Resources.NotFoundException ex) {
throw new FileNotFoundException("Resource does not exist: " + uri);
}
} else if (SCHEME_FILE.equals(scheme)) {
// Note: left here to avoid breaking compatibility. May be removed
// with sufficient testing.
return new FileInputStream(uri.getPath());
} else {
AssetFileDescriptor fd = openAssetFileDescriptor(uri, "r", null);
try {
return fd != null ? fd.createInputStream() : null;
} catch (IOException e) {
throw new FileNotFoundException("Unable to create stream");
}
}
}
this is the source code of ContentResolver, hope this can help you
Follow this URL . It will helps you
If you stored as in local drive.
String filePath = null;
Uri _uri = data.getData();
Log.d("","URI = "+_uri);
if(_uri!=null&&"content".equals(_uri.getScheme())){
Cursor cursor=this.getContentResolver().query(_uri,new String[]{android.provider.MediaStore.Images.ImageColumns.DATA},null,null,null);
cursor.moveToFirst();
filePath=cursor.getString(0);
cursor.close();
}else
{
filePath=_uri.getPath();
}
Related
I am trying to upload document from my app.
Everything working fine but when i choose file from drive.
data=Intent { act=android.intent.action.VIEW dat=content://com.google.android.apps.docs.storage.legacy/enc=ckpgt5KcEEF_JYniJQafRV_5pEnu_D5UAI1WF-Lu6h2Z_Vw4
(has extras) }}
Can any body know how to handle this file.
I had already handle all files and images only facing problem with google drive files.
I am getting this content://com.google.android.apps.docs.storage.legacy/enc=ckpgt5KcEEF_JYniJQafRV_5pEnu_D5UAI1WF-Lu6h2Z_Vw4 in intent data Uri.
Handle Uri received by Google-Drive files when selected through file chooser.
as stated earlier it receives Virtual File Uri.
I found this sample code simple and easy to understand.
the given code sample worked for me .hope it works in your case.
1.So detect this Uri is received by google drive.
public static File getFileFromUri(final Context context, final Uri uri) throws Exception {
if (isGoogleDrive(uri)) // check if file selected from google drive
{
return saveFileIntoExternalStorageByUri(context, uri);
}else
// do your other calculation for the other files and return that file
return null;
}
public static boolean isGoogleDrive(Uri uri)
{
return "com.google.android.apps.docs.storage.legacy".equals(uri.getAuthority());
}
2.if yes,the uri is stored to external path(here its root directory u can change it according to your need) and the file with that uri is created.
public static File saveFileIntoExternalStorageByUri(Context context, Uri uri) throws
Exception {
InputStream inputStream = context.getContentResolver().openInputStream(uri);
int originalSize = inputStream.available();
BufferedInputStream bis = null;
BufferedOutputStream bos = null;
String fileName = getFileName(context, uri);
File file = makeEmptyFileIntoExternalStorageWithTitle(fileName);
bis = new BufferedInputStream(inputStream);
bos = new BufferedOutputStream(new FileOutputStream(
file, false));
byte[] buf = new byte[originalSize];
bis.read(buf);
do {
bos.write(buf);
} while (bis.read(buf) != -1);
bos.flush();
bos.close();
bis.close();
return file;
}
public static String getFileName(Context context, Uri uri)
{
String result = null;
if (uri.getScheme().equals("content")) {
Cursor cursor = context.getContentResolver().query(uri, null, null, null, null);
try {
if (cursor != null && cursor.moveToFirst()) {
result = cursor.getString(cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
}
} finally {
cursor.close();
}
}
if (result == null) {
result = uri.getPath();
int cut = result.lastIndexOf('/');
if (cut != -1) {
result = result.substring(cut + 1);
}
}
return result;
}
public static File makeEmptyFileIntoExternalStorageWithTitle(String title) {
String root = Environment.getExternalStorageDirectory().getAbsolutePath();
return new File(root, title);
}
Note:Here the virtual file is retrieved from Intent getData() and used in context.getContentResolver().openInputStream(intent.getData()), this will return an InputStream. It's handle to get selected file from google drive.
for more info go through this link
I think you are getting Virtual File Uri from google drive
Read more about Virtual Files
FROM DOCS
Virtual Files
Android 7.0 adds the concept of virtual files to the Storage Access Framework. The virtual files feature allows your DocumentsProvider to return document URIs that can be used with an ACTION_VIEW intent even if they don't have a direct bytecode representation. Android 7.0 also allows you to provide alternate formats for user files, virtual or otherwise
Now question is how to check the the Uri is VirtualFile or not
You can find sample code from DOCS Open virtual files
first check that Uri is VirtualFile or not
private boolean isVirtualFile(Uri uri) {
if (!DocumentsContract.isDocumentUri(this, uri)) {
return false;
}
Cursor cursor = getContentResolver().query(
uri,
new String[] { DocumentsContract.Document.COLUMN_FLAGS },
null, null, null);
int flags = 0;
if (cursor.moveToFirst()) {
flags = cursor.getInt(0);
}
cursor.close();
return (flags & DocumentsContract.Document.FLAG_VIRTUAL_DOCUMENT) != 0;
}
The following code snippet shows how to check whether a virtual file can be represented as an image, and if so, gets an input stream from the virtual file
private InputStream getInputStreamForVirtualFile(Uri uri, String mimeTypeFilter)
throws IOException {
ContentResolver resolver = getContentResolver();
String[] openableMimeTypes = resolver.getStreamTypes(uri, mimeTypeFilter);
if (openableMimeTypes == null ||
openableMimeTypes.length < 1) {
throw new FileNotFoundException();
}
return resolver
.openTypedAssetFileDescriptor(uri, openableMimeTypes[0], null)
.createInputStream();
}
For more information of Virtual Files you can read below article
Virtual Files FAQ
Open files using storage access framework
An Android Storage Access Framework Example
I am trying to upload document from my app.
Everything working fine but when i choose file from drive.
data=Intent { act=android.intent.action.VIEW dat=content://com.google.android.apps.docs.storage.legacy/enc=ckpgt5KcEEF_JYniJQafRV_5pEnu_D5UAI1WF-Lu6h2Z_Vw4
(has extras) }}
Can any body know how to handle this file.
I had already handle all files and images only facing problem with google drive files.
I am getting this content://com.google.android.apps.docs.storage.legacy/enc=ckpgt5KcEEF_JYniJQafRV_5pEnu_D5UAI1WF-Lu6h2Z_Vw4 in intent data Uri.
Handle Uri received by Google-Drive files when selected through file chooser.
as stated earlier it receives Virtual File Uri.
I found this sample code simple and easy to understand.
the given code sample worked for me .hope it works in your case.
1.So detect this Uri is received by google drive.
public static File getFileFromUri(final Context context, final Uri uri) throws Exception {
if (isGoogleDrive(uri)) // check if file selected from google drive
{
return saveFileIntoExternalStorageByUri(context, uri);
}else
// do your other calculation for the other files and return that file
return null;
}
public static boolean isGoogleDrive(Uri uri)
{
return "com.google.android.apps.docs.storage.legacy".equals(uri.getAuthority());
}
2.if yes,the uri is stored to external path(here its root directory u can change it according to your need) and the file with that uri is created.
public static File saveFileIntoExternalStorageByUri(Context context, Uri uri) throws
Exception {
InputStream inputStream = context.getContentResolver().openInputStream(uri);
int originalSize = inputStream.available();
BufferedInputStream bis = null;
BufferedOutputStream bos = null;
String fileName = getFileName(context, uri);
File file = makeEmptyFileIntoExternalStorageWithTitle(fileName);
bis = new BufferedInputStream(inputStream);
bos = new BufferedOutputStream(new FileOutputStream(
file, false));
byte[] buf = new byte[originalSize];
bis.read(buf);
do {
bos.write(buf);
} while (bis.read(buf) != -1);
bos.flush();
bos.close();
bis.close();
return file;
}
public static String getFileName(Context context, Uri uri)
{
String result = null;
if (uri.getScheme().equals("content")) {
Cursor cursor = context.getContentResolver().query(uri, null, null, null, null);
try {
if (cursor != null && cursor.moveToFirst()) {
result = cursor.getString(cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
}
} finally {
cursor.close();
}
}
if (result == null) {
result = uri.getPath();
int cut = result.lastIndexOf('/');
if (cut != -1) {
result = result.substring(cut + 1);
}
}
return result;
}
public static File makeEmptyFileIntoExternalStorageWithTitle(String title) {
String root = Environment.getExternalStorageDirectory().getAbsolutePath();
return new File(root, title);
}
Note:Here the virtual file is retrieved from Intent getData() and used in context.getContentResolver().openInputStream(intent.getData()), this will return an InputStream. It's handle to get selected file from google drive.
for more info go through this link
I think you are getting Virtual File Uri from google drive
Read more about Virtual Files
FROM DOCS
Virtual Files
Android 7.0 adds the concept of virtual files to the Storage Access Framework. The virtual files feature allows your DocumentsProvider to return document URIs that can be used with an ACTION_VIEW intent even if they don't have a direct bytecode representation. Android 7.0 also allows you to provide alternate formats for user files, virtual or otherwise
Now question is how to check the the Uri is VirtualFile or not
You can find sample code from DOCS Open virtual files
first check that Uri is VirtualFile or not
private boolean isVirtualFile(Uri uri) {
if (!DocumentsContract.isDocumentUri(this, uri)) {
return false;
}
Cursor cursor = getContentResolver().query(
uri,
new String[] { DocumentsContract.Document.COLUMN_FLAGS },
null, null, null);
int flags = 0;
if (cursor.moveToFirst()) {
flags = cursor.getInt(0);
}
cursor.close();
return (flags & DocumentsContract.Document.FLAG_VIRTUAL_DOCUMENT) != 0;
}
The following code snippet shows how to check whether a virtual file can be represented as an image, and if so, gets an input stream from the virtual file
private InputStream getInputStreamForVirtualFile(Uri uri, String mimeTypeFilter)
throws IOException {
ContentResolver resolver = getContentResolver();
String[] openableMimeTypes = resolver.getStreamTypes(uri, mimeTypeFilter);
if (openableMimeTypes == null ||
openableMimeTypes.length < 1) {
throw new FileNotFoundException();
}
return resolver
.openTypedAssetFileDescriptor(uri, openableMimeTypes[0], null)
.createInputStream();
}
For more information of Virtual Files you can read below article
Virtual Files FAQ
Open files using storage access framework
An Android Storage Access Framework Example
In my app the user can select files and open them in the appropriate app (using an ACTION_VIEW intent).
I need to do some work on the data before giving it to the other app. So I'm using a streaming solution : I implemented a ContentProvider that implements openTypedAssetFile and writeDataToPipe (this method fills the output ParcelFileDescriptor created by openPipeHelper).
This works : I can open .pdf files, .txt files etc. The streaming seems correct.
I can open images usings 3-party apps.
However when I open an image using the Gallery, it doesn't work (Gallery shows a black screen), and I get the following exception :
fail to open myfile.jpg
UriImage(21890): java.io.FileNotFoundException: Not a whole file
I had a look at the Gallery source (here) and I could see that the exception is thrown here :
try {
if (MIME_TYPE_JPEG.equalsIgnoreCase(mContentType)) {
InputStream is = mApplication.getContentResolver()
.openInputStream(mUri);
mRotation = Exif.getOrientation(is);
Utils.closeSilently(is);
}
**mFileDescriptor = mApplication.getContentResolver()
.openFileDescriptor(mUri, "r");**
if (jc.isCancelled()) return STATE_INIT;
return STATE_DOWNLOADED;
} catch (FileNotFoundException e) {
Log.w(TAG, "fail to open: " + mUri, e);
return STATE_ERROR;
}
However, once in the Gallery app, if I select "Set as wallpaper", then I can see my image, it is then well streamed. So the problem appears when Gallery opens.
After a deeper look (in the ContentResolver code etc.) I couldn't understand why it behaves this way. It seems that Gallery does not support streaming files. Is that right ?
Do you have any idea ?
Thanks a lot.
String scheme = mUri.getScheme();
ContentResolver contentResolver = getContentResolver();
// If we are sent file://something or
// content://org.openintents.filemanager/mimetype/something...
if (scheme.equals("file")
|| (scheme.equals("content") && fileUri
.getEncodedAuthority().equals(
"org.openintents.filemanager"))) {
// Get the path
filePath = fileUri.getPath();
// Trim the path if necessary
// openintents filemanager returns
// content://org.openintents.filemanager/mimetype//mnt/sdcard/xxxx.jpg
if (filePath.startsWith("/mimetype/")) {
String trimmedFilePath = filePath
.substring("/mimetype/".length());
filePath = trimmedFilePath.substring(trimmedFilePath
.indexOf("/"));
}
} else if (scheme.equals("content")) {
// If we are given another content:// URI, look it up in the
// media provider
filePath = getFilePathFromContentUri(fileUri,
contentResolver);
} else {
Log.e("filePath--------->>>>>", filePath + "");
}
and
private String getFilePathFromContentUri(Uri selectedVideoUri,
ContentResolver contentResolver) {
String filePathString;
String[] filePathColumn = { MediaColumns.DATA };
Cursor cursor = contentResolver.query(selectedVideoUri, filePathColumn,
null, null, null);
cursor.moveToFirst();
int columnIndex = cursor.getColumnIndex(filePathColumn[0]);
filePathString = cursor.getString(columnIndex);
cursor.close();
return filePathString;
}
and now use this filepath ..
try {
if (MIME_TYPE_JPEG.equalsIgnoreCase(mContentType)) {
InputStream is = mApplication.getContentResolver()
.openInputStream(filePath); // ** change is here
mRotation = Exif.getOrientation(is);
Utils.closeSilently(is);
}
** mFileDescriptor = mApplication.getContentResolver()
.openFileDescriptor(filePath, "r"); **
if (jc.isCancelled()) return STATE_INIT;
return STATE_DOWNLOADED;
} catch (FileNotFoundException e) {
Log.w(TAG, "fail to open: " + mUri, e);
return STATE_ERROR;
}
this is work for me
When I pick the default ringtone, I get a Uri with path content://settings/system/ringtone. So there is no path ending with integer, as in returning a usual rintone (for example, content://media/internal/audio/media/38).
How can I now get a resource id from this default ringtone? Any parsing is impossible since there is no any integer identification.
If your check the setting db, you will find the content save to setting db is media uri. That means below uri actually is a map to media uri.
From:
content://settings/system/ringtone
To:
content://media/internal/audio/media/38
So you need to write a converter to do it:
public Uri uriMap(Uri uri) {
Uri mediaUri = uri;
if(uri.getAuthority().equals(Settings.AUTHORITY)) {
Cursor c = null;
try {
c = getContentResolver().query(uri,new String[]{
Settings.NameValueTable.VALUE},null,null,null);
if(c != null && c.moveToFirst()) {
String val = c.getString(0);
mediaUri = Uri.parse(val);
}
} catch (Exception e) {
}finally {
c.close();
}
}
Log.e(TAG,"" + uri + "->" + mediaUri);
return mediaUri;
}
I have registered my app to receive files (of any type, not just images) from other apps following this post.
I have implemented the solution that was answered but I cannot find a way to retrieve the "file name" of the data stream.
As an example from an Uri like:
content://downloads/all_downloads/5
I can get the stream out but I don't know anything about the name of the original file generating it.
Is there a way to retrieve it?
In MOST cases this will solve your problem:
Uri intentData = intent.getData();
if (intentData != null) {
String filePath;
if("content".equals(intent.getScheme()))
{
filePath = getFilePathFromContentUri(intentData);
}
else
{
filePath = intentData.getPath();
}
}
private String getFilePathFromContentUri(Uri selectedUri) {
String filePath;
String[] filePathColumn = {MediaColumns.DATA};
Cursor cursor = getContentResolver().query(selectedUri, filePathColumn, null, null, null);
cursor.moveToFirst();
int columnIndex = cursor.getColumnIndex(filePathColumn[0]);
filePath = cursor.getString(columnIndex);
cursor.close();
return filePath;
}
Is there a way to retrieve it?
Generally, no, because there may not be a name, in part because there may not be a file. You may be able to get an InputStream on the contents, but that does not mean that there is a file behind the InputStream.
There may be some specific hacks for some specific providers (e.g., MediaStore) to try to determine the file name associated with some data Uri, though such hacks may not be reliable.
onCreate()
Intent intent1 = getIntent();
String action = intent1.getAction();
String type = intent1.getType();
if (Intent.ACTION_SEND.equals(action) && type != null) {
this.handleSend(intent1);
}
void handleSend(Intent intent) {
try {
Uri imageUri = (Uri) intent.getParcelableExtra(Intent.EXTRA_STREAM);
imageShare.setImageURI(imageUri);
} catch (Exception e) {
e.printStackTrace();
}
}