i am implementing the functionality to save file in my app which is being sent by some exteral application.
i have provided support for single and mulitple files. Provided handling for all kind of files.
But i am not able to handle the following scenario.
I view a file from an email client -> View it in QuickOffice -> Click on send -> Choose my app->Then click on save in my app.
In that i get the path in following wrapped in the exception
java.io.FileNotFoundException: /file:/data/data/com.qo.android.sp.oem/files/temp/Error.log: open failed: ENOENT (No such file or directory)
I have seen this post which is quite useful for handling uri which has content scheme
Get filename and path from URI from mediastore
Below is my code
Uri uri = (Uri) iterator.next();
if ("content".equals(uri.getScheme())) {
filePath = getFilePathFromContentUri(uri, hostAcitvity.getContentResolver());
}
else {
filePath = uri.getPath();
}
fileName = uri.getLastPathSegment();
fileSize = hostAcitvity.getContentResolver().openInputStream(uri).available();
Code for getFilePathFromContentUri
private String getFilePathFromContentUri(Uri selectedVideoUri, ContentResolver contentResolver)
{
String filePath;
String[] filePathColumn = { MediaColumns.DATA };
Cursor cursor = contentResolver.query(selectedVideoUri, filePathColumn, null, null, null);
cursor.moveToFirst();
int columnIndex = cursor.getColumnIndex(filePathColumn[0]);
filePath = cursor.getString(columnIndex);
cursor.close();
return filePath;
}
Then i wrap the path in a FileInputStream which is throwing the above exception
Not able to resolve the file path properly. Is this the correct way of finding the path ?
cheers,
Saurav
I have seen this post which is quite useful for handling uri which has content scheme
That never worked reliably and will work even less reliably in the future.
Is this the correct way of finding the path ?
No, because there is no requirement that every Uri map to a path on a filesystem that you can access.
Use getInputStream() on ContentResolver to get an InputStream on the Uri, and consume the data that way.
Related
I am testing app on nexus 5 which has marshmallow version of android. I am looked into several method to get file path from uri but all the time it returns null. This method works for me on jelly bean and also on kitkat but not on marshmallow.
public String getPath(Uri uri) {
Log.d(TAG,"Uri GET PATH "+uri);
String[] projection = {MediaStore.Images.Media.DATA};
Cursor cursor = context.getContentResolver().query(uri, projection, null, null, null);
Log.d(TAG,"Cursor value "+cursor);
int Column_Index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
Log.d(TAG,"Column Index "+Column_Index);
cursor.moveToFirst();
String ImagePath = cursor.getString(Column_Index);
Log.d(TAG,"ImagePath of "+ImagePath);
cursor.close();
return ImagePath;
}
I also tried this this and also so many others but getting file path from uri is always null.
Uri return from Intent:content://com.android.providers.media.documents/document/image%3A2304`
I am looked into several method to get file path from uri but all the time it returns null
There is no file path. 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 and a FileOutputStream on some file that you control (e.g., in getCacheDir()) to copy the content to the file, then use the resulting file.
I have a method below:
private String getRealPathFromUriForVideos(Uri selectedVideoUri) {
String wholeID = DocumentsContract.getDocumentId(selectedVideoUri);
String id = wholeID.split(":")[1];
String[] column = { MediaStore.Video.Media.DATA };
String sel = MediaStore.Video.Media._ID + "=?";
Cursor cursor = getContentResolver().query(MediaStore.Video.Media.EXTERNAL_CONTENT_URI, column, sel, new String[]{ id }, null);
String filePath = "";
int columnIndex = cursor.getColumnIndex(column[0]);
if (cursor.moveToFirst()) {
filePath = cursor.getString(columnIndex);
}
cursor.close();
return filePath;
}
This works just fine getting the file for videos that hte user selects. However, I want to allow users to also create new videos (from my app) and then get the URI and the file from there. The URI for newly created videos is: content://media/external/video/media/41. For selected videos is like content://com.android.providers.media.documents/document/video%3A42.
It works with the second one but not the first one. First one I get IllegalArgumentException because its not a document URI. How can I get the file from the first URI?
This works just fine getting the file for videos that hte user selects
It may work in a few situations. It will not work in general. A Uri that you get from something like ACTION_OPEN_DOCUMENT does not have to represent a file, let alone one that you can access via the filesystem, let alone one that this script-kiddie algorithm will let you access.
The URI for newly created videos is: content://media/external/video/media/41
Not necessarily. I suppose that there is a way that you get a Uri like that for a recorded video, though off the top of my head I cannot think of a recommended way that would give you such a Uri. If you are using MediaRecorder or ACTION_VIDEO_CAPTURE, you create your own file (and, for ACTION_VIDEO_CAPTURE, your own Uri for that file). And, if you are creating your own file, you know where that file is.
Need to be able to upload to my server
For the video, record to a file that you control, then use that file.
Use some library that lets you upload from a Uri or InputStream. Otherwise:
Use ContentResolver and openFileInput() to get an InputStream on the content represented by the Uri
Create a FileOutputStream on some file that you control (e.g., in getCacheDir())
Copy the content from the InputStream to the OutputStream
Use your copy for the upload
Delete your copy when the work is done
You treat a foreign Uri as if it were a URL to a Web server: stream the content.
Seems to get it from the second URI I need this:
private String getRealPathFromUriForImagesAndVideo(Uri contentUri) {
Cursor cursor = null;
try {
String[] proj = {MediaStore.Images.Media.DATA};
cursor = getContentResolver().query(contentUri, proj, null, null, null);
int column_index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
cursor.moveToFirst();
return cursor.getString(column_index);
} catch (Exception e) {
return contentUri.getPath();
} finally {
if (cursor != null) {
cursor.close();
}
}
}
I want to get file path from Uri for a video. The following method works fine when testing with a real device, however, it fails (returns null) when testing on emulator.
public String getRealPathFromURI(Context context, Uri contentUri) {
Cursor cursor = null;
try {
String[] proj = {MediaStore.Video.Media.DATA};
cursor = context.getContentResolver().query(contentUri, proj, null, null, null);
int column_index = cursor.getColumnIndexOrThrow(MediaStore.Video.Media.DATA);
cursor.moveToFirst();
return cursor.getString(column_index);
} catch (Exception e) {
e.printStackTrace();
} finally {
if (cursor != null) {
cursor.close();
}
}
return null;
}
What is the correct way of getting file path from uri on emulator?
The following method works fine when testing with a real device
Only on the the device that you tried, and only for the app that you tried. Particularly on Android 4.4+, your approach will be unreliable. That is because a Uri is not a file. On older versions of Android, for a Uri from the MediaStore, your approach might work.
Nowadays, do not attempt to get a file for a Uri. Consume the Uri as you are supposed to, using methods on ContentResolver to get an InputStream, the MIME type, etc.
What is the correct way of getting file path from uri on emulator?
There is none. There does not have to be a file path associated with a Uri, let alone a path that your app is able to access using Java file I/O.
As CommonsWare mentioned, an Uri is NOT a File. The general way to deal with Uri is to use an inputstream and save the content as a file (assuming that's what you are looking for). What i typically do is
get the metadata associated with the Uri (to get title / type of data / size)
get the content via an input stream to save it on the device as a file.
Take a look at the "Examine document metadata" and "get an inputstream" on this page: https://developer.android.com/guide/topics/providers/document-provider.html
What I am trying to achieve is sounds very familiar, it has been posted many times here and there in Stack Overflow as well, but I'm unable to get it done.
The scenario is, I receive a mail with attachment having custom extension in it. The extension is recognized by my app and it needs the FilePath to process it.
Currently, when I get the attachment in my app using getIntent().getData() all I get is path of the form content://
I have seen methods to convert media content of the type content:// to FilePath like /sdcard/file.ext but I was unable to convert the attachment using that. May be its obvious.
Is there any way that I can process the content:// type without actually downloading it.
Currently from the k9 mail app, when I get the custom extension, it shows my app in the list and opens it through it, but I need FilePath like /sdcard/file.ext and I'm only able to get content:// type.
I hope I made the question clear.
Please Help.
Regards.
A content:// Uri does not necessarily point to a file on the sdcard.
It is more likely that it points to any kind of data stored in a database
or to a content provider that gives you access to the private file storage of another app.
I think the later one is the case with mail attachments (if the content provider is not requesting it directly from a web server). So converting the content:// Uri to a path will not work.
I did the following (not sure if it works also for k9 mail app)
Uri uri = intent.getData();
if (uri.getScheme().equals("content")) {
String fileName = ContentProviderUtils.getAttachmentName(this, uri);
if (fileName.toLowerCase().endsWith(".ext")) {
InputStream is = this.getContentResolver().openInputStream(uri);
// do something
} else {
// not correct extension
return;
}
} else if (uri.getScheme().equals("file")) {
String path = uri.getPath();
if (path.toLowerCase().endsWith(".ext")) {
InputStream is = new FileInputStream(path);
// do something
} else {
// not correct extension
return;
}
}
The attachment name can be found by
public static String getAttachmentName(Context ctxt, Uri contentUri) {
Cursor cursor = ctxt.getContentResolver().query(contentUri, new String[]{MediaStore.MediaColumns.DISPLAY_NAME}, null, null, null);
String res = "";
if (cursor != null){
cursor.moveToFirst();
int nameIdx = cursor.getColumnIndex(MediaStore.MediaColumns.DISPLAY_NAME);
res = cursor.getString(nameIdx);
cursor.close();
}
return res;
}
I am desperately trying to send a captured video to a server. The problem is that the URI that is given by the built-in camera application is not the real file path. It looks like this - /content:/media/external/video/media/19.
How can I access the real path or the data directly from this kind of URIs?
After reading the android documentation I saw that it looks like a content provider's notation, but I still don't have a clue how to reach the data that I need. Please help!!!
thanks in advance
How can I access the real path or the data directly from this kind of URIs?
You don't. It might not exist as a file. Or, it might not exist as a file that you can read except via the ContentProvider.
Instead, use a ContentResolver to open an InputStream on that Uri, and use that InputStream to transfer the data to a server.
public String getRealPathFromURI(Context context, Uri contentUri) {
Cursor cursor = null;
try {
String[] proj = { MediaStore.Images.Media.DATA };
cursor = context.getContentResolver().query(contentUri, proj, null, null, null);
int column_index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
cursor.moveToFirst();
return cursor.getString(column_index);
} finally {
if (cursor != null) {
cursor.close();
}
}
}
see following post for
Get filename and path from URI from mediastore