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Editorial Team
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Asked: 2026-05-16T02:55:38+00:00

I have a servlet coded in Scala. I have some code like this in

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I have a servlet coded in Scala. I have some code like this in there:

def message = <HTML><HEAD><TITLE>Test</TITLE></HEAD><BODY>{value}</BODY></HTML>
def value = "Hello <BR/> World"

The corresponding HTML code generated for value is 

Hello &lt;BR/&gt; World

How do I get it to generate the HTML code (shown below)?

Hello <BR/> World

Thanks in advance

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1 Answer

  1. Editorial Team

    If you don’t mind changing the type of value to xml.Elem, you can do

    def value = <xml:group>Hello <BR/> World</xml:group>

    Addition

    In my opinion, you should type as much XML inline as possible. Only then will you have compile time validation of the input. All other solutions either give you a runtime exception at some point (say, you forgot some /) or might even break your XML layout.

    However, if you really want to have an easy transform, you could do this:

    class XmlString(str: String) {
  def assumeXML = xml.XML.loadString("<xml:group>" + str + "</xml:group>")
  def toUnparsedXML = new xml.Unparsed(str)
}
implicit def stringToXmlString(str: String) = new XmlString(str)

def value = "Hello <BR/> World"

    and then (for some reason, it still shows the <xml:group> part; you could get rid of it with xml.NodeSeq.fromSeq(value.assumeXML.child) or similar)

    def message = <HTML><HEAD><TITLE>Test</TITLE></HEAD><BODY>{value.assumeXML}</BODY></HTML>

    or even (well, you would not need the implicit here, Unparsed(value) would do)

    def message = <HTML><HEAD><TITLE>Test</TITLE></HEAD><BODY>{value.toUnparsedXML}</BODY></HTML>

    The assumeXML method will fail with at runtime, if you provide invalid XML; toUnparsedXML will accept all input, even data that is potentially dangerous.

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