Android: Getting a file URI from a content URI

androidfileuri

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.

Best 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)

Related Solutions

Python – How to check whether a file exists without exceptions

If the reason you're checking is so you can do something like if file_exists: open_it(), it's safer to use a try around the attempt to open it. Checking and then opening risks the file being deleted or moved or something between when you check and when you try to open it.

If you're not planning to open the file immediately, you can use os.path.isfile

Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.

import os.path
os.path.isfile(fname)

if you need to be sure it's a file.

Starting with Python 3.4, the pathlib module offers an object-oriented approach (backported to pathlib2 in Python 2.7):

from pathlib import Path

my_file = Path("/path/to/file")
if my_file.is_file():
    # file exists

To check a directory, do:

if my_file.is_dir():
    # directory exists

To check whether a Path object exists independently of whether is it a file or directory, use exists():

if my_file.exists():
    # path exists

You can also use resolve(strict=True) in a try block:

try:
    my_abs_path = my_file.resolve(strict=True)
except FileNotFoundError:
    # doesn't exist
else:
    # exists

Python – How to copy a file in Python

shutil has many methods you can use. One of which is:

from shutil import copyfile
copyfile(src, dst)

# 2nd option
copy(src, dst)  # dst can be a folder; use copy2() to preserve timestamp

Copy the contents of the file named src to a file named dst. Both src and dst need to be the entire filename of the files, including path.
The destination location must be writable; otherwise, an IOError exception will be raised.
If dst already exists, it will be replaced.
Special files such as character or block devices and pipes cannot be copied with this function.
With copy, src and dst are path names given as strs.

Another shutil method to look at is shutil.copy2(). It's similar but preserves more metadata (e.g. time stamps).

If you use os.path operations, use copy rather than copyfile. copyfile will only accept strings.

Best Answer

Related Solutions

Python – How to check whether a file exists without exceptions

Python – How to copy a file in Python

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