How to eval a string val in Scala?

Question

I have scala expression stored in String variable:

val myExpr = "(xml \ \"node\")"

How do I execute this?

s"${myExpr}"

Right now it only gives me the string contents

What I'm trying to achieve is parsing user string input in the form:

"/some/node/in/xml"

and get that corresponding node in Scala:

(xml \ "node" \ "in" \ "xml")

What do you want to achieve with this in the end? I'm confident there's a better task for the job. — nietaki
– nietaki, Commented Jun 4, 2015 at 22:02
Made edits to my post to describe what I'm doing. I need to execute user generated xPath expressions. — Sean Bollin
– Sean Bollin, Commented Jun 4, 2015 at 22:05

som-snytt · Accepted Answer · 2015-06-04 22:24:12Z

1

For the REPL, my init includes:

implicit class interpoleter(val sc: StringContext) {def i(args: Any*) = $intp interpret sc.s(args: _*) }

with which

scala> val myExpr = "(xml \\ \"node\")"
myExpr: String = (xml \ "node")

scala> val xml = <x><node/></x>
xml: scala.xml.Elem = <x><node/></x>

scala> i"${myExpr}"
res3: scala.xml.NodeSeq = NodeSeq(<node/>)
res2: scala.tools.nsc.interpreter.IR.Result = Success

because isn't code really just a string, like everything else?

answered Jun 4, 2015 at 22:24

som-snytt

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Comments

om-nom-nom · Accepted Answer · 2015-06-04 22:05:18Z

0

Probably, there is some more idiomatic way in recent scala versions, but you can use Twitter's Eval for that:

val i: Int = new Eval()("1 + 1") // => 2

answered Jun 4, 2015 at 22:05

om-nom-nom

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2 Comments

Travis Brown Over a year ago

Please don't use Eval! We're trying to kill it!

om-nom-nom Over a year ago

@TravisBrown what's the reasoning behind this? Can you elaborate on this?

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