How to add a file to an Android project, deploy it

2019-02-17 04:45发布

I have an Android (2.2) project in Eclipse (Helios). I want to add an MP3 file to the project so that the MP3 file is deployed to the device along with the application.

I then would like to open the file as a File object, which means I'd need to know the full path (?) to the file on the device, but I don't know how paths are specified in Android.

2条回答
一夜七次
2楼-- · 2019-02-17 05:04

Apparently there is a bug in Froyo that prevents WAV playback.
Audio files should be placed in the "res/raw" directory of your project. Then use the id to play it (or attempt to play it)

MediaPlayer mp = MediaPlayer.create(context, R.raw.sound_file_1);
mp.start();

Info: http://developer.android.com/guide/topics/media/index.html
Example (mp3): http://www.helloandroid.com/tutorials/musicdroid-audio-player-part-i

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可以哭但决不认输i
3楼-- · 2019-02-17 05:22

Ok, I saw this on the source of another projet, so I didn't really come up with it, but it works.

To add any file to a project, and later be able to use it, you need to put the file (binary, xml, or whatever) on the assets folder of your project.

In this example I will just copy the asset to the filesystem, so I can later access it as any other user file. You can access the assets directly too, take a look at Resources on the documentation.

public void copyfile(String fileName)
      {
        if (!new File(fileName).exists()){
            try
            {
              InputStream localInputStream = getAssets().open(fileName);
              FileOutputStream localFileOutputStream = getBaseContext().openFileOutput(fileName, MODE_PRIVATE);

              byte[] arrayOfByte = new byte[1024];
              int offset;
              while ((offset = localInputStream.read(arrayOfByte))>0)
              {
                localFileOutputStream.write(arrayOfByte, 0, offset);
              }
              localFileOutputStream.close();
              localInputStream.close();
              // The next 3 lines are because I'm copying a binary that I plan
              // to execute, so I need it to have execute permission.
              StringBuilder command = new StringBuilder("chmod 700 ");
              command.append(basedir + "/" + paramString);
              Runtime.getRuntime().exec(command.toString());
              Log.d(TAG, "File " + paramString + " copied successfully.");
            }
            catch (IOException localIOException)
            {
                localIOException.printStackTrace();
                return;
            }
        }
        else
            Log.d(TAG, "No need to copy file " + paramString);
      }

I believe there is probably a better way to copy the file, but this one works, and does not slow down my app even if it's called from onCreate (all files I copy are below 100kb though, so for bigger files, you probably want a separate thread)

Here is how to get the path to your files:

String path = getBaseContext().getFilesDir().getAbsolutePath();

If you want to write the file to another path, say "/sdcard/DCIM/appPictures", I believe you can use this code:

FileOutputStream outFile = FileOutputStream("/sdcard/DCIM/appPictures/" + fileName);

and then copy it byte by byte like in the example above.

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