How to open a file in raw folder in Android

问题

I am usin MultipartEntity and I am trying to refer to the file in the raw folder. Here is the code:

MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart(new FormBodyPart("file", new FileBody(new File("test.txt"))));

The test.txt file is in my res/raw folder. When I execute the code I get the following exception : FileNotFoundException: /test.txt: open failed: ENOENT (No such file or directory)

Can anyone help me with this?

回答1:

Unfortunately you can not create a File object directly from the raw folder. You need to copy it or in your sdcard or inside the application`s cache.

you can retrieve the InputStream for your file this way

    InputStream in = getResources().openRawResource(R.raw.yourfile);

  try {
       int count = 0;
       byte[] bytes = new byte[32768];
       StringBuilder builder = new StringBuilder();
       while ( (count = in.read(bytes,0, 32768)) > 0) {
           builder.append(new String(bytes, 0, count));
       }

       in.close();
       reqEntity.addPart(new FormBodyPart("file", new StringBody(builder.toString())));
   } catch (IOException e) {
       e.printStackTrace();
   }

回答2:

You can put the file in the /res/raw directory, where the file will be indexed and is accessible by an id in the R file:

InputStream is = getResources().openRawResource(R.raw.test);
System.out.println(is);

来源：https://stackoverflow.com/questions/16741433/how-to-open-a-file-in-raw-folder-in-android