How to read a resource file to a byte array in scala?

Question

I just want to read file in resource and get a byte array? Can someone help me with that?

Almost a duplicate of http://stackoverflow.com/questions/27360977/how-to-read-files-from-resources-folder-in-scala?rq=1 (that one tries to read lines instead of bytes, but maybe you can adapt it, for example by combining it with http://stackoverflow.com/questions/7598135/how-to-read-a-file-as-a-byte-array-in-scala) — Thilo, Nov 17 '15 at 11:19

poki2 · Answer 1 · 2020-09-01T07:43:28.993

3

Note: this requires Java 9+

I use the following to read a file in my resources as byte array:

getClass.getResourceAsStream("/file-in-resource-folder").readAllBytes()

edited Sep 01 '20 at 07:43

answered Aug 31 '20 at 14:23

poki2

71
5

score 2 · Answer 2 · edited May 23 '17 at 12:15

2

As described in How to read a file as a byte array in Scala, the following fragment should do the trick:

def slurp(resource: String) = {
  val bis = new BufferedInputStream(getClass.getResource(resource))
  try Stream.continually(bis.read).takeWhile(-1 !=).map(_.toByte).toArray
  finally bis.close()
}

edited May 23 '17 at 12:15

Community

1
1

answered Nov 17 '15 at 12:01

Norwæ

1,575
8
18

This approach is very slow and heavy on CPU, because it reads the file byte by byte, one byte at a time. – m.bemowski Sep 27 '18 at 14:28

How to read a resource file to a byte array in scala?

2 Answers2