Use path of source in a scala script file

Question

I am looking for a way to use the full path of the *.scala file that is executed by

scala /path/to/file/file.scala

The reason is that within my script I would like to use paths, which are relative to the location where file.scala is saved. So, say I want to call /path/to/file/file_second.scala from inside of file.scala while calling file.scala by

scala ./to./file/file.scala

as the directory I invoked scala from is /path (i.e. what System.getProperty("user.dir") would return). All I know priori is that file_second.scala is in the same directory as file.scala without actually knowing the full path.

I have tried. among others, getClass.getResource("").split(":")(1), but this will either return /path/./ or /tmp/scala.tmp/

I guess there must be a clean solution as scala -savecompiled will create a *.jar at exactly that directory I want to work on, but I just cannot figure out how... :/

Many thanks for any help.

I would like to add a solution I developed from the second answer (when I read it weeks ago, I haven't read answer Nr.1 yet) for scala scripts invoked on linux systems. It is the way I use to do it now. Thanks for all the answers, I learned sooo much from it! I just don't know which answer to flag as "accepted". The solution: — Wayne Jhukie
– Wayne Jhukie, Commented Jun 4, 2012 at 20:06
#! /bin/bash INSTALL_DIR="$( cd "$( dirname "BASH_SOURCE[0]" )" && pwd )" exec scala -savecompiled -DPARAM_INSTALL_DIR=$INSTALL_DIR "$0" "$@" !# val install_dir = System.getProperty("PARAM_INSTALL_DIR") + java.io.File.separator — Wayne Jhukie
– Wayne Jhukie, Commented Jun 4, 2012 at 20:08

Jamie Eisenhart · Accepted Answer · 2013-06-26 19:22:13Z

7

Script preamble:

#!/bin/sh
export SCRIPT_PATH=$(readlink -f "$0") && exec scala "$0" "$@"
!#

Read the value with:

val scriptPath = System.getenv("SCRIPT_PATH")

answered Jun 26, 2013 at 19:22

Jamie Eisenhart

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2 Comments

Jonathan Leonard Over a year ago

This is the best answer (and as usual it is the last).

F. P. Freely Over a year ago

export SCRIPT_DIR="$(dirname "$(readlink -f "$0")")" && exec scala "$0" "$@"

Connor Doyle · Accepted Answer · 2012-04-05 03:48:19Z

1

I've done this in the past in Java. It's not pretty, and it may not be 100% reliable depending on the details of your class loader situation.

Here is some Scala code that prints out the location of its own class file.

import java.net.URL

object LocationTest extends App {
  val classDirURL = LocationTest.getClass.getResource("LocationTest.class")
  val classDirPath = classDirURL.getPath.replaceAll("%20", " ")
  println(classDirPath)
}

prints

/Users/connor/desktop/LocationTest.class

edited Apr 5, 2012 at 3:48

answered Apr 5, 2012 at 2:33

Connor Doyle

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Comments

philwalk · Accepted Answer · 2012-04-27 13:14:09Z

The following script will print it's filesystem path. Three different approaches are tried in succession, until one of them succeeds.

import java.io.File

printf("[%s]\n",new File(getScriptName).getCanonicalPath)
System.exit(0)

def getScriptName:String = {
  // this approach works on the sun jvm:
  val jcmd = System.getProperty("sun.java.command","")
  jcmd.split("""\s+""").toList.reverse.head match {
  case name if new File(name).exists => name
  case _ =>
    // for this to work, you need to set -Dscala.script.name at script launch
    System.getProperty("scala.script.name","") match {
    case name if new File(name).exists => name
    case _ =>
      // last resort: derive script basename from stackdump
      val basename = {
        val filenames = new RuntimeException("").getStackTrace.map { it
          => it.getFileName
        }
        filenames.indexOf("NativeMethodAccessorImpl.java") match {
          case -1 =>  filenames.reverse.head
          case num => filenames(num - 1)
        }
      }
      // then search the system PATH for the script file
      import scala.collection.JavaConversions._
      val sep = System.getProperty("path.separator")
      val path = System.getenv("PATH").split(sep)
      path.map(new File(_,basename)).find( _.exists ) match {
        case Some(scriptfile) => scriptfile.getCanonicalPath
        case _ => basename // better than nothing
      }
    }
  }
}

The easiest and most portable is the 2nd approach, which depends on use of a wrapper script, as in the following example, for defining Property 'scala.script.name' before calling $SCALA_HOME/bin/scala. The wrapper script needs to be in your PATH ahead of $SCALA_HOME/bin, of course. The result is more bulletproof, but is only applicable to scripts.

#!/usr/bin/env sh
for SCRIPT_NAME; do true ; done # set SCRIPT_NAME to the last argument
JAVA_OPTS="$JAVA_OPTS -Dscala.script.name=${SCRIPT_NAME}"
$SCALA_HOME/bin/scala "$@"

Use of this wrapper around the scala launcher simplifies usage to the following:

#!/usr/bin/env scala
!#
printf("%s\n",System.getProperty("scala.script.name"))

user unknown · Accepted Answer · 2012-04-05 00:20:44Z

You can invoke your program with a param:

scala -DPARAM_DIR=tmp Foo

and check for the param inside the program:

object Foo extends App {
  val dir = System.getProperty ("PARAM_DIR")
  println ("Dir: " + dir) 
}

So this wouldn't (necessarily) use the same directory where Foo.scala is located, but it would be flexible, and can be told at runtime where to look for a class.

In normal cases you either bundle classes in a Jar, or you set a common classpath, and then you might have a package which is resolved relatively to the classpath:

For a class to.file.File in

 /path/to/file/file.scala

it would be the classpath /path, for a class File in a package file it would be /path/to/.

If your classpath is /path/to/file, the second class should be reachable simply by name.

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