Get Xpath from command line in scala

Question

import scala.io.Source
import scala.xml.{Node, NodeSeq, XML}

object scalaTest extends App {


var test= <name>person
<fname>doe
<first>john</first>
</fname>
</name>


var s=scala.io.StdIn.readLine
var result=s.asInstanceOf[NodeSeq]

println(result)


}

When i provide xpath test\"fname"\"first" on command line, it throws,

test\"fname"\"first"
Exception in thread "main" java.lang.ClassCastException: java.lang.String cannot be cast to scala.xml.NodeSeq
    at scalaTest$.delayedEndpoint$scalaTest$1(scalaTest.scala:15)
    at scalaTest$delayedInit$body.apply(scalaTest.scala:4)
    at scala.Function0.apply$mcV$sp(Function0.scala:34)
    at scala.Function0.apply$mcV$sp$(Function0.scala:34)
    at scala.runtime.AbstractFunction0.apply$mcV$sp(AbstractFunction0.scala:12)
    at scala.App.$anonfun$main$1$adapted(App.scala:76)
    at scala.collection.immutable.List.foreach(List.scala:389)
    at scala.App.main(App.scala:76)
    at scala.App.main$(App.scala:74)
    at scalaTest$.main(scalaTest.scala:4)
    at scalaTest.main(scalaTest.scala)

How to provide xpath as test\"fname"\"first" in command line and get result as,

<first>john</first>

I have no experience with XML in Scala, but shouldn't s be somehow relatated to test in your code? Or how are these related? — user unknown
– user unknown, Commented Mar 9, 2018 at 12:33

Andrey Tyukin · Accepted Answer · 2018-03-09 14:15:52Z

1

You cannot just cast a String into a NodeSeq. You could attempt to parse it with XML.loadString(...). But this is also not what you want, because your string does not contain xml, it contains some xpath-like selection. I don't know whether Scala's built-in XML actually supports Xpath. I only know that it has a few funny methods on NodeSeqs like \ and \\ that look somewhat like Xpath. What you could do is:

get input string from stdin
split your input string at the '\'-character
foldLeft the test-xml node using the \-method

Something like this works:

import scala.xml.NodeSeq
val test = <name>person<fname>doe<first>john</first></fname></name>
val xpath = """fname\first""" // or read from StdIn
val selectedElement = xpath.split("\\\\").foldLeft[NodeSeq](test){
  case (elem, tagName) => elem \ tagName 
}

// gives result:
// res20: scala.xml.NodeSeq = NodeSeq(<first>john</first>)

Or shorter:

val selectedElement = xpath.split("\\\\").foldLeft[NodeSeq](test)(_ \ _)

Another note: you of course cannot use the String "test" in your StdIn input. How is the compiler supposed to know that it has anything to do with the local test-variable?

edited Mar 9, 2018 at 14:15

answered Mar 9, 2018 at 13:59

Andrey Tyukin

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3 Comments

s.Doe Over a year ago

This works fine! But when i need to give XPATH some thing like these (test\"fname"\"first")(0), I do not think native scala XML doesn't support this!

Andrey Tyukin Over a year ago

@s.Doe I think Scala's native support for XML was more popular in 2010 than it is now. For example here, Prof. Odersky himself says that "XML literals {...} stick{s} out like a sore thumb". I don't know about scala's XML support that much, but maybe it would be easier to use a separate library that takes XPath much more seriously.

s.Doe Over a year ago

Thanks a lot @Andrey

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