Android Getting a file URI from a content URI

Question

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

User · Answer

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)

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

Inspired answers are Jason LaBrun  amp  Darth Raven  Trying already answered approaches led me to below solution which may mostly cover cursor null cases  amp  conversion from content     to file      To convert file  read amp 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  amp  amp  fileCursor moveToFirst              int cIndex   fileCursor getColumnIndex OpenableColumns DISPLAY NAME           if  cIndex    -1                fileName   fileCursor getString cIndex                       return fileName

User · Answer

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  amp  amp   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

User · Answer

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    quot temp file quot    if  fileExtension    null   quot   fileExtension quot  else  quot  quot          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    gt  0            target write buf  0  length

User · Answer

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  gt   Build VERSION CODES KITKAT          DocumentProvider     if  isKitKat  amp  amp  DocumentsContract isDocumentUri context  uri             System out println  quot getPath   uri   quot    uri toString             System out println  quot getPath   uri authority   quot    uri getAuthority             System out println  quot getPath   uri path   quot    uri getPath                 ExternalStorageProvider         if   quot com android externalstorage documents quot  equals uri getAuthority                   final String docId   DocumentsContract getDocumentId uri               final String   split   docId split  quot   quot                final String type   split 0               System out println  quot getPath   docId   quot    docId    quot   split   quot    split length    quot   type   quot    type                   This is for checking Main Memory             if   quot primary quot  equalsIgnoreCase type                     if  split length  gt  1                        return Environment getExternalStorageDirectory      quot   quot    split 1     quot   quot                     else                       return Environment getExternalStorageDirectory      quot   quot                                        This is for checking SD Card               else                   return  quot storage quot     quot   quot    docId replace  quot   quot    quot   quot                                       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

[android] Android: Getting a file URI from a content URI?

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