Scala XML transformation with Rewrite rules - scala

Question

I have a XML template with some fields predefined in it. I want to build new XML based on it template with new Value values using RewriteRules.
ex.
template:
val template = <xml>
<Persons>
<Name>Persons</Name>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value></Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value></Value>
</LastName>
</Person>
</Persons>
</xml>
case class Person(fullName: String, lastName: String)
val persons = Seq(Person("John Smith", "Smith"), Person("Bob Saver", "Saver"))
output should be:
<xml>
<Persons>
<Name>Persons</Name>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value>John Smith</Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value>Smith</Value>
</LastName>
</Person>
<Person>
<FullName>
<Name>Name of the field</Name>
<Value>Bob Saver</Value>
</FullName>
<LastName>
<Name>Name of the field</Name>
<Value>Saver</Value>
</LastName>
</Person>
</Persons>
</xml>
Is it possible to do with RewriteRules?

Answer 1

You do not need RewriteRules for this purposes. You can define variables in your xml template.
scala> def template(id: String = "defaultValue can be here") = <someXml>{id}</someXml>
template: (id: String)scala.xml.Elem
scala> template("person")
res4: scala.xml.Elem = <someXml>person</someXml>
scala> template("person2")
res5: scala.xml.Elem = <someXml>person2</someXml>
otherwise
Scala - replace xml element with specific text