Scala Process - Capture Standard Out and Exit Code

Question

I\'m working with the Scala scala.sys.process library.

I know that I can capture the exit code with ! and the output with !!

Answer 1

The one-line-ish use of BasicIO or ProcessLogger is appealing.

scala> val sb = new StringBuffer
sb: StringBuffer = 

scala> ("/bin/ls /tmp" run BasicIO(false, sb, None)).exitValue
res0: Int = 0

scala> sb
res1: StringBuffer = ...

or

scala> import collection.mutable.ListBuffer
import collection.mutable.ListBuffer

scala> val b = ListBuffer[String]()
b: scala.collection.mutable.ListBuffer[String] = ListBuffer()

scala> ("/bin/ls /tmp" run ProcessLogger(b append _)).exitValue
res4: Int = 0

scala> b mkString "\n"
res5: String = ...

Depending on what you mean by capture, perhaps you're interested in output unless the exit code is nonzero. In that case, handle the exception.

scala> val re = "Nonzero exit value: (\\d+)".r.unanchored
re: scala.util.matching.UnanchoredRegex = Nonzero exit value: (\d+)

scala> Try ("./bomb.sh" !!) match {
     | case Failure(f) => f.getMessage match {
     |   case re(x) => println(s"Bad exit $x")
     | }
     | case Success(s) => println(s)
     | }
warning: there were 1 feature warning(s); re-run with -feature for details
Bad exit 3