How to properly decompress gz archive in Scala

Question

I a newbie in Scala and I have small task that requires me to decompress *.gz file from resources directory. So I want a proper way to do that to be able to parse file content after. Surely I have read some articles in the past, like: ONE TWO THREE

I can parse file content that is not archieved but cannot handle gz archive right now. Looks like I'm missing something small as I am newbie in Java and in Scala too.

Scala version - 2.21.0

I have a part of my code below:

object ResourceLoader {
    def loadResource(fileName: String): Try[InputStream] = Try(getClass.getResourceAsStream(fileName))


    def loadResource(fileName: String): Try[List[String]] =
        for {
            resourceStream <- loadResource(fileName)
            resourceContent = Source.fromInputStream(resourceStream).getLines.toList
        } yield resourceContent
}

Then I can iterate over not archieved file like:

        val content = ResourceLoader.loadResourceContent("/test_text.csv") recover {
            case e: FileNotFoundException => println(s"Requested file not found: $e")
            case e: SecurityException => println(s"Permission denied: $e")
            case e: Exception => println(s"An unknown exception occurred: $e")
        }
        content.foreach(println)

But cannot understand how to decompress gz archive first and then iterate over it.

I expect to use GZIPInputStream instead of getResourceAsStream in loadResource function but can't understand how to do that in proper way.

Thank you in advance for any help!

Can not test right now, but maybe something like this will work: `Source.fromInputStream(new GZIPInputStream(Thread.currentThread.getContextClassLoader.getResourceAsStream(resource)))`. — Luis Miguel Mejía Suárez, May 10 '19 at 16:30

score 4 · Accepted Answer · answered May 13 '19 at 09:13

As @Luis commented, this is what you can do:


val inputStream = Thread.currentThread().getContextClassLoader.getResourceAsStream("test_text.csv.gz")
val gzipFileSource: BufferedSource = Source.fromInputStream(new GZIPInputStream(inputStream))

println(gzipFileSource.getLines.toList.head)

How to properly decompress gz archive in Scala

1 Answers1