How to read file from res/raw by name

Question

I want to open a file from the folder res/raw/. I am absolutely sure that the file exists. To open the file I have tried

File ddd = new File("res/raw/example.png");

The command

ddd.exists();

yields FALSE. So this method does not work.

Trying

MyContext.getAssets().open("example.png");

ends up in an exception with getMessage() "null".

Simply using

R.raw.example

is not possible because the filename is only known during runtime as a string.

Why is it so difficult to access a file in the folder /res/raw/ ?

possible duplicate of [Dynamic Resource Loading Android](http://stackoverflow.com/questions/3648942/dynamic-resource-loading-android) — Simon, Apr 09 '13 at 21:27
Why is it only known at runtime as a string? Could you keep a mapping between the strings and the IDs? — kabuko, Apr 09 '13 at 21:28
Read the docs: http://developer.android.com/guide/topics/resources/providing-resources.html — dmon, Apr 09 '13 at 21:30
Thank you for the links. I will give a correct solution as a separate answer. — Bernd, Apr 10 '13 at 19:13

score 176 · Accepted Answer · edited Jan 29 '16 at 19:31

176

With the help of the given links I was able to solve the problem myself. The correct way is to get the resource ID with

getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",
                             "raw", getPackageName());

To get it as a InputStream

InputStream ins = getResources().openRawResource(
            getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",
            "raw", getPackageName()));

edited Jan 29 '16 at 19:31

quietmint

13,885
6
48
73

answered Apr 10 '13 at 19:18

Bernd

3,405
3
18
21

18

Just to clarify: Do **not** prepend `raw/` for the file name. Valid example: `.getIdentifier("products","raw", getPackageName());` for `res/raw/products.json` – JJD Jan 24 '14 at 11:16
5

As stated by JJD, you don't need the raw/ before the FILENAME_WITHOUT_EXTENSION. I found that if you wish to use the extension, for example I have two files abc.jpg and abc.zip, I will add them both to raw folder as abc_jpg and abc_zip (extensions in raw folder are ignored), then assign String variable filename = abc.jpg or abc.zip, then use InputStream ins = getResources().openRawResource(getResources().getIdentifier(filename.replace(".", "_"), "raw", getPackageName())); Hope this helps someone :) – Bruce Jun 28 '14 at 02:38
remember to close any streams you open – Pierre Aug 15 '17 at 20:03
i need the filepath... how can i get it instead of the inputstream? – Rafael Lima May 27 '18 at 00:29
1

It is not worked for localized raw folders. Any Idea?? – Rushabh Shah Jul 30 '21 at 10:54

score 50 · Answer 2 · answered Apr 09 '13 at 21:27

Here is example of taking XML file from raw folder:

 InputStream XmlFileInputStream = getResources().openRawResource(R.raw.taskslists5items); // getting XML

Then you can:

 String sxml = readTextFile(XmlFileInputStream);

when:

 public String readTextFile(InputStream inputStream) {
        ByteArrayOutputStream outputStream = new ByteArrayOutputStream();

        byte buf[] = new byte[1024];
        int len;
        try {
            while ((len = inputStream.read(buf)) != -1) {
                outputStream.write(buf, 0, len);
            }
            outputStream.close();
            inputStream.close();
        } catch (IOException e) {

        }
        return outputStream.toString();
    }

The problem is that I can't use R.raw.XYZ because the name of XYZ is only available as a string. But the link of simon help me solve the problem. I will give a correct solution as a separate answer. — Bernd, Apr 10 '13 at 19:14
this works perfectly fine `InputStream input = Context.getResources().openRawResource(R.raw.configjson);` — Someone Somewhere, Dec 12 '13 at 18:40

score 14 · Answer 3 · edited Jul 11 '13 at 22:53

14

You can read files in raw/res using getResources().openRawResource(R.raw.myfilename).

BUT there is an IDE limitation that the file name you use can only contain lower case alphanumeric characters and dot. So file names like XYZ.txt or my_data.bin will not be listed in R.

edited Jul 11 '13 at 22:53

Avi

21,182
26
82
121

answered Jul 11 '13 at 22:35

user2574545

141
1
2

10

underscore is also allowed in resource file names – laalto Jul 11 '13 at 22:55
3

Also numbers are allowed (not as the first character). – Phantômaxx Oct 28 '14 at 08:14
Not as the first character...because that would not be a valid Java variable identifier – Peter Chaula Sep 26 '19 at 06:29

score 4 · Answer 4 · answered Feb 01 '19 at 07:26

Here are two approaches you can read raw resources using Kotlin.

You can get it by getting the resource id. Or, you can use string identifier in which you can programmatically change the filename with incrementation.

Cheers mate

// R.raw.data_post

this.context.resources.openRawResource(R.raw.data_post)
this.context.resources.getIdentifier("data_post", "raw", this.context.packageName)

How to read file from res/raw by name

4 Answers4

Linked

Related