CODEWITHSUNDEEP
androidfileuri
JMRboosties
JMRboosties

Reputation: 15740

Android: Getting a file URI from a content URI?

In my app the user is to select an audio file which the app then handles. The problem is that in order for the app to do what I want it to do with the audio files, I need the URI to be in file format. When I use Android's native music player to browse for the audio file in the app, the URI is a content URI, which looks like this:

content://media/external/audio/media/710

However, using the popular file manager application Astro, I get the following:

file:///sdcard/media/audio/ringtones/GetupGetOut.mp3

The latter is much more accessible for me to work with, but of course I want the app to have functionality with the audio file the user chooses regardless of the program they use to browse their collection. So my question is, is there a way to convert the content:// style URI into a file:// URI? Otherwise, what would you recommend for me to solve this problem? Here is the code which calls up the chooser, for reference:

Intent ringIntent = new Intent();
ringIntent.setType("audio/mp3");
ringIntent.setAction(Intent.ACTION_GET_CONTENT);
ringIntent.addCategory(Intent.CATEGORY_OPENABLE);
startActivityForResult(Intent.createChooser(ringIntent, "Select Ringtone"), SELECT_RINGTONE);

I do the following with the content URI:

m_ringerPath = m_ringtoneUri.getPath();
File file = new File(m_ringerPath);

Then do some FileInputStream stuff with said file.

Upvotes: 194

Views: 218724

Answers (11)

zhonglaoban
zhonglaoban

Reputation: 41

private String getNativePath(String javaString) {
    Uri uri = Uri.parse(javaString);
    ContentResolver resolver = this.getApplicationContext().getContentResolver();
    ParcelFileDescriptor pfd = null;
    try {
        pfd = resolver.openFileDescriptor(uri, "r");
    } catch (FileNotFoundException e) {
        throw new RuntimeException(e);
    }
    return "pipe:" + pfd.detachFd();
};

you can get the fd and return a fd string to native code

Upvotes: 0

solru
solru

Reputation: 151

If you're absolutely sure that your content: Uri points to a real file, you can do the following trick:

context.contentResolver.openFileDescriptor(uri, "r").use { fileDescriptor ->
    if (fileDescriptor == null) {
        throw NullPointerException("ParcelFileDescriptor from $uri was null")
    }
    val path = "/proc/${android.os.Process.myPid()}/fd/${fileDescriptor.fd}"
    val fileUri = Uri.fromFile(File(path).canonicalFile)
}

Here we open the file using ContentResolver, construct a path to an underlying Linux file descriptor and resolve a real file path by getting canonicalFile.

Upvotes: 0

Mahesh Mahi
Mahesh Mahi

Reputation: 41

You can use the following android package which can be easier a bit for you

https://github.com/Blankj/AndroidUtilCode

Using the above package the code can be like

To Import use below Line

import com.blankj.utilcode.util.UriUtils;

Your code can be like

File f = UriUtils.uri2File(result);

Thanks

Upvotes: 1

Thracian
Thracian

Reputation: 66674

If you have a content Uri with content://com.externalstorage... you can use this method to get absolute path of a folder or file on Android 19 or above.

public static String getPath(final Context context, final Uri uri) {
    final boolean isKitKat = Build.VERSION.SDK_INT >= Build.VERSION_CODES.KITKAT;

    // DocumentProvider
    if (isKitKat && DocumentsContract.isDocumentUri(context, uri)) {
        System.out.println("getPath() uri: " + uri.toString());
        System.out.println("getPath() uri authority: " + uri.getAuthority());
        System.out.println("getPath() uri path: " + uri.getPath());

        // ExternalStorageProvider
        if ("com.android.externalstorage.documents".equals(uri.getAuthority())) {
            final String docId = DocumentsContract.getDocumentId(uri);
            final String[] split = docId.split(":");
            final String type = split[0];
            System.out.println("getPath() docId: " + docId + ", split: " + split.length + ", type: " + type);

            // This is for checking Main Memory
            if ("primary".equalsIgnoreCase(type)) {
                if (split.length > 1) {
                    return Environment.getExternalStorageDirectory() + "/" + split[1] + "/";
                } else {
                    return Environment.getExternalStorageDirectory() + "/";
                }
                // This is for checking SD Card
            } else {
                return "storage" + "/" + docId.replace(":", "/");
            }

        }
    }
    return null;
}

You can check each part of Uri using println. Returned values for my SD card and device main memory are listed below. You can access and delete if file is on memory, but I wasn't able to delete file from SD card using this method, only read or opened image using this absolute path. If you find a solution to delete using this method, please share.

SD CARD

getPath() uri: content://com.android.externalstorage.documents/tree/612E-B7BF%3A/document/612E-B7BF%3A
getPath() uri authority: com.android.externalstorage.documents
getPath() uri path: /tree/612E-B7BF:/document/612E-B7BF:
getPath() docId: 612E-B7BF:, split: 1, type: 612E-B7BF

MAIN MEMORY

getPath() uri: content://com.android.externalstorage.documents/tree/primary%3A/document/primary%3A
getPath() uri authority: com.android.externalstorage.documents
getPath() uri path: /tree/primary:/document/primary:
getPath() docId: primary:, split: 1, type: primary

If you wish to get Uri with file:/// after getting path use

DocumentFile documentFile = DocumentFile.fromFile(new File(path));
documentFile.getUri() // will return a Uri with file Uri

Upvotes: 14

andermirik
andermirik

Reputation: 50

you can get filename by uri with simple way

Retrieving file information

fun get_filename_by_uri(uri : Uri) : String{
    contentResolver.query(uri, null, null, null, null).use { cursor ->
        cursor?.let {
            val nameIndex = it.getColumnIndex(OpenableColumns.DISPLAY_NAME)
            it.moveToFirst()
            return it.getString(nameIndex)
        }
    }
    return ""
}

and easy to read it by using

contentResolver.openInputStream(uri)

Upvotes: -2

Shaon
Shaon

Reputation: 2724

Try this....

get File from a content uri

fun fileFromContentUri(context: Context, contentUri: Uri): File {
    // Preparing Temp file name
    val fileExtension = getFileExtension(context, contentUri)
    val fileName = "temp_file" + if (fileExtension != null) ".$fileExtension" else ""

    // Creating Temp file
    val tempFile = File(context.cacheDir, fileName)
    tempFile.createNewFile()

    try {
        val oStream = FileOutputStream(tempFile)
        val inputStream = context.contentResolver.openInputStream(contentUri)

        inputStream?.let {
            copy(inputStream, oStream)
        }

        oStream.flush()
    } catch (e: Exception) {
        e.printStackTrace()
    }

    return tempFile
}

private fun getFileExtension(context: Context, uri: Uri): String? {
    val fileType: String? = context.contentResolver.getType(uri)
    return MimeTypeMap.getSingleton().getExtensionFromMimeType(fileType)
}

@Throws(IOException::class)
private fun copy(source: InputStream, target: OutputStream) {
    val buf = ByteArray(8192)
    var length: Int
    while (source.read(buf).also { length = it } > 0) {
        target.write(buf, 0, length)
    }
}

Upvotes: 30

Rafael Nobre
Rafael Nobre

Reputation: 5131

This is an old answer with deprecated and hacky way of overcoming some specific content resolver pain points. Take it with some huge grains of salt and use the proper openInputStream API if at all possible.

You can use the Content Resolver to get a file:// path from the content:// URI:

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();
}
Log.d("","Chosen path = "+ filePath);

Upvotes: 53

Muhammed Yalçın Kuru
Muhammed Yalçın Kuru

Reputation: 668

Inspired answers are Jason LaBrun & Darth Raven. Trying already answered approaches led me to below solution which may mostly cover cursor null cases & conversion from content:// to file://

To convert file, read&write the file from gained uri

public static Uri getFilePathFromUri(Uri uri) throws IOException {
    String fileName = getFileName(uri);
    File file = new File(myContext.getExternalCacheDir(), fileName);
    file.createNewFile();
    try (OutputStream outputStream = new FileOutputStream(file);
         InputStream inputStream = myContext.getContentResolver().openInputStream(uri)) {
        FileUtil.copyStream(inputStream, outputStream); //Simply reads input to output stream
        outputStream.flush();
    }
    return Uri.fromFile(file);
}

To get filename use, it will cover cursor null case 

public static String getFileName(Uri uri) {
    String fileName = getFileNameFromCursor(uri);
    if (fileName == null) {
        String fileExtension = getFileExtension(uri);
        fileName = "temp_file" + (fileExtension != null ? "." + fileExtension : "");
    } else if (!fileName.contains(".")) {
        String fileExtension = getFileExtension(uri);
        fileName = fileName + "." + fileExtension;
    }
    return fileName;
}

There is good option to converting from mime type to file extention

 public static String getFileExtension(Uri uri) {
    String fileType = myContext.getContentResolver().getType(uri);
    return MimeTypeMap.getSingleton().getExtensionFromMimeType(fileType);
}

Cursor to obtain name of file

public static String getFileNameFromCursor(Uri uri) {
    Cursor fileCursor = myContext.getContentResolver().query(uri, new String[]{OpenableColumns.DISPLAY_NAME}, null, null, null);
    String fileName = null;
    if (fileCursor != null && fileCursor.moveToFirst()) {
        int cIndex = fileCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
        if (cIndex != -1) {
            fileName = fileCursor.getString(cIndex);
        }
    }
    return fileName;
}

Upvotes: 14

Umar Ata
Umar Ata

Reputation: 4258

Well I am bit late to answer,but my code is tested

check scheme from uri:

 byte[] videoBytes;

if (uri.getScheme().equals("content")){
        InputStream iStream =   context.getContentResolver().openInputStream(uri);
            videoBytes = getBytes(iStream);
        }else{
            File file = new File(uri.getPath());
            FileInputStream fileInputStream = new FileInputStream(file);     
            videoBytes = getBytes(fileInputStream);
        }

In the above answer I converted the video uri to bytes array , but that's not related to question, I just copied my full code to show the usage of FileInputStream and InputStream as both are working same in my code.

I used the variable context which is getActivity() in my Fragment and in Activity it simply be ActivityName.this

context=getActivity(); //in Fragment

context=ActivityName.this;// in activity

Upvotes: 4

Darth Raven
Darth Raven

Reputation: 189

Trying to handle the URI with content:// scheme by calling ContentResolver.query()is not a good solution. On HTC Desire running 4.2.2 you could get NULL as a query result.

Why not to use ContentResolver instead? https://stackoverflow.com/a/29141800/3205334

Upvotes: 9

Jason LeBrun
Jason LeBrun

Reputation: 13293

Just use getContentResolver().openInputStream(uri) to get an InputStream from a URI.

http://developer.android.com/reference/android/content/ContentResolver.html#openInputStream(android.net.Uri)

Upvotes: 195

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