How to read from zipped xml files in Scala code?

Question

How do I access XML data files directly from a zipped file in my Scala program? Are there any direct ways to programmatically unzip and read contents in my Scala code?

Parthenogenesis

Answer 1

Here are a couple of ways of doing it in 2.8.1:

cat > root.xml << EOF
<ROOT>
<id>123</id>
</ROOT>
EOF
zip root root.xml

and then in the REPL:

val rootzip = new java.util.zip.ZipFile("root.zip")
import collection.JavaConverters._
val entries = rootzip.entries.asScala
entries foreach { e =>
    val x = scala.xml.XML.load(rootzip.getInputStream(e))
    println(x)
}

or something like:

val rootzip = new java.util.zip.ZipFile("root.zip")
import scala.collection.JavaConversions._
rootzip.entries.
  filter (_.getName.endsWith(".xml")).
  foreach { e => println(scala.xml.XML.load(rootzip.getInputStream(e))) }

Tyro

Thanks a lot. This one really helped the most. I pasted an implicit method code for converting the Java Enumeration to Scala list. collection.JavaConverters._ and asScala() helped reduce the code complexity. A lot of really helpful examples for both XML and ZIP file reading in scala. Thanks a ton. – Parthenogenesis

Answer 2

You can use the Java package java.util.zip: http://download.oracle.com/javase/6/docs/api/java/util/zip/package-summary.html

Metallic

Which does have gzip (though not tar.gz) support, as the OP's tag requests. – Rectangular

I see no 'tar' here, just 'gzip'. :) A GZIPInputStream should be just what the doctor ordered. Or, if it's actually a PKZIP file, something else in the same package will work (with an extra helping of accidental complexity) – Jamilla

hmmmm...any quick sample code to look at? Am jut being a little lazy, thats it. – Parthenogenesis

Answer 3

I personally prefer TrueZip. It allows you to treat archive files as a virtual file system, providing the same interface as standard Java file I/O.

Corenda