scala.xml.pull, but it's deprecated, for good reasons.) I'd suggest looking into what Java XML libraries support streaming, and use one of those from Scala
Hello, @all I would like to write an xml Elem to a file, but I want to set the line separator between each node of the elem, so when opening the file under windows or ubuntu I have the correct format. for example :
val lineSep = System.getProperty("line.separator")
val xmlData : Elem = <person>
<firstName>John</firstName>
<lastName>Doe</lastName>
<emails>
<email type=”primary”>john.doe@noone.com</email>
<email type=”secondary”>john.doe@noone.com</email>
</emails>
<address>
<street>595 Market Street</street>
<city>San Francisco</city>
<zip>94105</zip>
</address>
</person>How to write it into a file and considering the lineSep?
I see the following change in behavior between scala 2.12 and 2.13:
scala 2.12:
scala> import scala.xml._
import scala.xml._
scala> val a: NodeSeq = <a>Hello</a>
a: scala.xml.NodeSeq = <a>Hello</a>
scala> a ++ <b>hi</b>
res0: scala.xml.NodeSeq = NodeSeq(<a>Hello</a>, <b>hi</b>)Scala 2.13:
scala> a ++ <b>hi</b>
res0: Seq[scala.xml.Node] = List(<a>Hello</a>, <b>hi</b>)That change has a very deep reach for us, because Lift doesn't interpret both type in the same way: it applies template transformation one time for a NodeSeq (it's only one template), and N times for List[Node] (it's a list of template). Our web application is totally broken, and it's almost impossible to find all places which would need a type ascription (and it would bloat a lot the resulting code.
Is there a way to get back the old behavior? Perhaps we are missing a new implicit? Perhaps the implicit precedence in NodeSeq changed and it could be adpated?
Any insight would be appreciated.
Hi guys, i have a question for you. I'm trying to map Elem children but the result is not what i expected.
val elem: Elem =
<a>
<b>1</b>
<b>2</b>
<b>3</b>
</a>
val result = elem.child.map(k => <c>{k.text}</c>)Returns
ArrayBuffer(
<c></c>,
<c>1</c>,
<c></c>,
<c>2</c>,
<c></c>,
<c>3</c>,
<c></c>
)Can someone explain me why this happend ? why empty nodes ?
elem.child.map {
case e: Elem => <c>k.text</c>
case n => n
}
res0: Seq[scala.xml.Node] = List( , <c>k.text</c>, , <c>k.text</c>, , <c>k.text</c>, )
encoding setting , but that won't help. The XML declaration could have an encoding attribute, or not, but I don't think that's tracked either. Parts of the code look like there was an effort to try and capture the encoding, but it was abandoned and never completed.
"org.scala-lang.modules" %% "scala-xml" % "2.0.0-M5" was found, however packages in scala-xml do not contain any classes. tried both 2.0.0-RC1 and 2.0.0-M5 versions.
Is xml string interpolation implemented as a feature of Scala 3 ? If positive, how can I make
xml"""
|<p>Hello World!</p>
|""".stripMargincompile? I always receive a value xml is not a member of StringContext error message.
1 reply
Hi there,
Parse XML using Scala with colon (:) inside XML tag. is this possible?
val xml =
"""
|<saml2p:Response Version="2.0" xmlns:saml2p="urn:oasis:names:tc:SAML:2.0:protocol">
| <saml2:Issuer xmlns:saml2="urn:oasis:names:tc:SAML:2.0:assertion">
| https://metadata
| </saml2:Issuer>
|</saml2p:Response>
|""".stripMargin
val doc = scala.xml.XML.loadString(xml)
val node = (doc \ "Issuer")
println(node)
println(node \@("saml2"))
println(doc \@("xmlns"))xmlns:saml2, but I don't get result :(
\ operator only works by element label, and not prefix, and the same for \@.
scala> doc.label
val res0: String = Response
scala> doc.prefix
val res1: String = saml2p
scala> doc.attributes
val res2: scala.xml.MetaData = Version="2.0"
scala> doc.namespace
val res3: String = urn:oasis:names:tc:SAML:2.0:protocol
scala> doc \@ "Version"
val res4: String = 2.0
scala> val nodes = (doc \ "Issuer")
val nodes: scala.xml.NodeSeq = ...
scala> nodes.map(elem => (elem.namespace, elem.label, elem.text.trim))
val res5: Seq[(String, String, String)] = List((urn:oasis:names:tc:SAML:2.0:assertion,Issuer,https://metadata))