CODEWITHSUNDEEP
scala
Rjj
Rjj

Reputation: 297

Scala how to create file to array

I have a file as below,

AB*xyz*1234~
CD*mny*769~
MN*bvd*2345~
AB*zar*987~

here i would like to split the data using(*) and line always ends with ~.

code.

val bufferedSource = Source.fromFile(sample.txt)
    var array = Array[String]()
    for (line <- bufferedSource.getLines()) {
      array = line.split("\\~")
      val splittedRow=array.split("\\*")

      }

I want output as below which should be first words from all lined,

Array(AB,CD,MN,AB).

Upvotes: 0

Views: 477

Answers (3)

Puneeth Reddy V
Puneeth Reddy V

Reputation: 1568

You can also use collect to get what you want

val s = Source.fromFile("fileName").getLines().collect{
  case e => e.split("\\*").apply(0)
}.toArray
//s: Array[String] = Array(AB, CD, MN, AB)

Upvotes: 0

Timothy Jones
Timothy Jones

Reputation: 22125

This is probably better achieved by mapping over the lines from the file, and selecting the part of the string you want to keep:

val array = 
  Source.fromFile(sample.txt)  // Open the file
    .getLines()                // get each line
    .toArray                   // as an array
    .map(s => s.split("\\*").head)  // Then select just the first part
                                    // of the line, split by * characters

Upvotes: 1

Ramesh Maharjan
Ramesh Maharjan

Reputation: 41957

You can simply do 

Source.fromFile(filename).getLines().map(line => line.split("\\*").head).toArray
//res0: Array[String] = Array(AB, CD, MN, AB)

Upvotes: 1

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