Is there anyway to implement optional command line arguments in Scala?

Question

I want to implement command lines in my program so that they can be optional but as of now the only way I can get my program to run is if I input 3 arguments when I run the program in the command line. Is there any way to have it so I only have to input a minimum of one argument and have the other 2 as optional?

Here is the start of my code:

 import scala.collection.mutable.ListBuffer
 import util.control.Breaks._
 object part3 {
   var newStartTime = 0
   var newEndTime = 0
   var csvStartTime = 0
   var csvEndTime = 0
   def main(args: Array[String]): Unit = {
     var csV = args(0)
     var daY = args(1)
     var time1 = args(2)
     var time2 = args(3)
     newEndTime = checkLengths(time1, time2)
     val bufferedSource = io.Source.fromFile(csV)
     var z = new ListBuffer[String]()
     var z2 = new ListBuffer[String]()
     var i = 0

Updated code:

val fileName = args(0)
courseCode1 = args(1)
val courseCode2: Option[String] = Try(args(2)).toOption
val courseCode3: Option[String] = Try(args(3)).toOption

val extraCourseSelection = for {
    c2 <- courseCode2
    c3 <- courseCode3
}
courseCodeFixed1 = courseCode1.toUpperCase.patch(4, " ", 0)
if(c2!=null)
{
    courseCodeFixed2 = c2.toUpperCase.patch(4, " ", 0)
}
if(c3!=null)
{
    courseCodeFixed3 = c3.toUpperCase.patch(4, " ", 0)
}
print(courseCodeFixed2)
print(courseCodeFixed3)

I tried implementing this solution but now it does not want to read any of my command line arguments at all

Your code has too much information (why show us the variables z, z2, and i?), not enough information (what's the default value for daY?), and confusing information (you say you want 1 minimum and 2 optional arguments but you code requires 4 arguments). It also doesn't compile. Code posted to SO should be minimal, complete, and reproducible. — jwvh
– jwvh, Commented Dec 3, 2019 at 2:34

jwvh · Accepted Answer · 2019-12-05 04:12:42Z

2

val fileName    = args(0)
val courseCode1 = args(1).toUpperCase.patch(4, " ", 0)
val courseCode2 = args.lift(2).map(_.toUpperCase.patch(4, " ", 0))
val courseCode3 = args.lift(3).map(_.toUpperCase.patch(4, " ", 0))

You can add .getOrElse() to provide a default value if you don't want Option[String] results.

answered Dec 5, 2019 at 4:12

jwvh

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1 Comment

Richard Desouza Over a year ago

There ya go I'm new to this stackoverflow thing lol. Thanks again!

Brandon Mott · Accepted Answer · 2019-12-03 02:24:00Z

1

You can use scala.util.Try to catch the index out of bounds exception and use toOption to convert Try[_] to Option[_]:

import scala.util.Try

...

val csV: Option[String] = Try(args(0)).toOption
val daY: Option[String] = Try(args(1)).toOption
val time1: Option[String] = Try(args(2)).toOption
val time2: Option[String] = Try(args(3)).toOption

Then to use the optional values you can use a for comprehension:

val newEndTime = for {
  t1 <- time1
  t2 <- time2
} yield checkLengths(t1, t2)

answered Dec 3, 2019 at 2:24

Brandon Mott

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2 Comments

jwvh Over a year ago

Try with .toOption is overkill. Just use lift: args.lift(0)

Richard Desouza Over a year ago

I tried implementing this solution but when I pass arguments now the program just doesn't read them at all?(including the first argument) Please see my updated code above

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