CODEWITHSUNDEEP
scalainputstreamscala-streams
Alexandre Annic
Alexandre Annic

Reputation: 10738

Why Source.fromInputStream is working with GZIPInputStream but not with ZipInputStream

I am trying to stream a zip file.

The following chunk of code prints line by line as expected:

val inputStream = new GZIPInputStream(new FileInputStream("/some/path"))
val source = Source.fromInputStream(inputStream)
for(line <- source.getLines) {
  println(line)
}

But this one doesn't do anyting (it doesn't even exit):

val inputStream = new ZipInputStream(new FileInputStream("/some/path"))
val source = Source.fromInputStream(inputStream)
for(line <- source.getLines) {
  println(line)
}

The only difference is the usage of GZIPInputStream instead of ZipInputStream. Both class implements InputStream.

Am I missing something ? Or is there any workaroud ?

Upvotes: 0

Views: 435

Answers (1)

Dima
Dima

Reputation: 40500

Gzip is just a compressed file, that can be de-compressed on the fly as you are reading from Source. Zip isn't really a stream, it's just one of many java misnomers (take a look at the interface), it's more like a directory, containing several files, that you can traverse via ZipEntry, and read each one separately via Source. There is no real content at the top level, just a directory listing, so no "lines" to fetch via Source.

In a nutshell, you just iterate through entries, creating a new Source for each one. Something like this:

   Iterator
     .continually(zip.getNextEntry)
     .takeWhile(_ != null)
     .map { e => 
        e.getName -> Source.fromInputStream(zip).getLines.toList
     }

(This creates a Map of name of each file in zip to its entire content n memory, probably, not what you want at all, just an illustration of what you can do to access that content via Source)

Upvotes: 4

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