Scala: read user input as single string

Question

I need get all user input in stdin as single string in scala 2.12 (supposed the data would copy-pasted by single action), something like this:

please copy data:
word1
word2
word3

And I need get string with following data:

val str = "word1\nword2\nword3"

my current approach is not working, just hanging forever:

import scala.collection.JavaConverters._
val scanner: Iterator[String] = new Scanner(System.in).asScala

val sb = new StringBuilder
while (scanner.hasNext) {
  sb.append(scanner.next())
}
val str = sb.toString()

Although this can print the input:

import scala.collection.JavaConverters._
val scanner: Iterator[String] = new Scanner(System.in).asScala

scanner foreach println

I'm looking for idiomatic way of doing the job

Mario Galic · Accepted Answer · 2019-07-28T16:10:19.607

2

Try

LazyList
  .continually(StdIn.readLine())
  .takeWhile(_ != null)
  .mkString("\n")

as inspired by https://stackoverflow.com/a/18924749/5205022

On my machine I could terminate the input with ^D.

In Scala 2.12 replace LazyList with Stream.

edited Jul 28 '19 at 16:10

answered Jul 28 '19 at 15:47

Mario Galic

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thanks for the answer, but this doesn't work for me properly. Btw what is LazyList? I used Iterator.continually instead but it is hanging forever unless I introduced special ending character – Normal Jul 28 '19 at 16:01
ohh I realised it is a new collection from scala 2.13, does it make difference vs Iterator ? – Normal Jul 28 '19 at 16:02
2

@Normal In Scala 2.12, just replace `LazyList` with `Stream`. In what way is it not working properly, that is, what error are you getting? – Mario Galic Jul 28 '19 at 16:09

Scala: read user input as single string

1 Answers1