Question

I am parsing a series of XML responses from a external data store. During which I must test for the existence of a child node and – if it exists – test its value. To achieve that I have the following code:

...
  val properties = for {
    val row <- root \\ "ResultDescription"
    val cond:Boolean = checkDetectionNode(row) match {
      case Some(nodeseq) => {
          val txt = nodeseq.text.toLowerCase
          if (txt contains "non-detect")
            false
          else
            true
      }
      case None => true
    }
    if (cond)
    val name = (row \ "CharacteristicName").text
    if (charNameList.exists(s => s == name) == false)
  } yield {
    getObservedProperty(name) match {
      case Some(property) => {
          charNameList = name :: charNameList
          property
      }
    }
  }
...

checkDetectionNode is defined as such:

private def checkDetectionNode(row: scala.xml.NodeSeq) : Option[scala.xml.NodeSeq] = {
  if ((row \ "ResultDetectionConditionText") != null)
    Some[scala.xml.NodeSeq]((row \ "ResultDetectionConditionText"))
  else
    None
}

The above code results in an unspecified error of “illegal start of simple expression” on the val name... line. To be honest I am not a Scala programmer or even a functional programmer (always was more partial to OO/imperative). I’ve only been using Scala for a few days and been basing most of what I know from Java and lambda operators. Unfortunately, I don’t really have the time to sit down and really learn Scala like I wish I could. Deadlines, make fools of us all.

I am hoping that someone could take a look and let me know if there is something I am doing wrong (as I am sure there is). I tried to limit the code shown to, what I hope, is relevant to the question. However, please let me know if any additional code is needed.

Thanks

Answer 1

The xml is a distraction here. The problem is the if (cond) at the end is not acting as a guard, it just looks like it should, the compiler thinks is the start of a new if ‘then’ part.

In the following example:

val l = List(1,2,3,4,5)

val r = for {
    i <- l 
      if (i > 2)
    x <- Some(i * 2)
  } yield x

you’ll get List(6,8,10) as you might expect.

Using

val r = for {
    val i <- l if (i > 2)
    val x <- Some(i * 2)
  } yield x

should get you a deprecation warning, and

val r = for {
    val i <- l 
    if (i > 2)
    val x <- Some(i * 2)
  } yield x

gets

error: illegal start of simple expression
  val x <- Some(i * 2)

Simply remove the val in front of your Generator (Pattern1 ‘<-’ Expr [Guard]), and you can resume normal service. It also flows a bit more nicely without the vals in the for loop I find.