Scala Cookbook

14.4. Running a Shell Command from the REPL

Problem

You want to be able to run a shell command from within the Scala REPL, such as listing the files in the current directory.

Solution

Run the command using the :sh REPL command, then print the output. The following example shows how to run the Unix ls -al command from within the REPL, and then show the results of the command:

scala> :sh ls -al
res0: scala.tools.nsc.interpreter.ProcessResult = `ls -al` (6 lines, exit 0)

scala> res0.show
total 24
drwxr-xr-x   5 Al  staff  170 Jul 14 17:14 .
drwxr-xr-x  29 Al  staff  986 Jul 14 15:27 ..
-rw-r--r--   1 Al  staff  108 Jul 14 15:34 finance.csv
-rw-r--r--   1 Al  staff  469 Jul 14 15:38 process.scala
-rw-r--r--   1 Al  staff  412 Jul 14 16:24 process2.scala

Alternatively you can import the scala.sys.process package, and then use the normal Process and ProcessBuilder commands described in Recipe 12.10:

scala> import sys.process._
import sys.process._

scala> "ls -al" !
total 24
drwxr-xr-x   5 Al  staff  170 Jul 14 17:14 .
drwxr-xr-x  29 Al  staff  986 Jul 14 15:27 ..
-rw-r--r--   1 Al  staff  108 Jul 14 15:34 finance.csv
-rw-r--r--   1 Al  staff  469 Jul 14 15:38 process.scala
-rw-r--r--   1 Al  staff  412 Jul 14 16:24 process2.scala
res0: Int = 0

Scala’s -i option

Although those examples show the correct approach, you can improve the situation by loading your own custom code when you start the Scala interpreter. For instance, I always start the REPL in my /Users/Al/tmp directory, and I have a file in that directory named repl-commands ...