substituting xml values programmatically with scala

Question

I'm writing a tool to update some xml files (pom.xml in this case) with scala because the effort it would take in java is significantly higher than (in theory) it is with scala. I can parse the xml file just fine, but I need to replace nodes in the existing xml and rewrite the result. for example:

<dependency>
    <groupId>foo</groupId>
    <artifactId>bar</artifactId>
    <version>1.0-SNAPSHOT</version>
</dependency>

So I want to find all nodes like this and replace them with:

<dependency>
    <groupId>foo</groupId>
    <artifactId>bar</artifactId>
    <version>1.0</version> <!-- notice the lack of -SNAPSHOT here -->
</dependency>

So, I can get all the version nodes simply enough, but how to replace them with the node that I want?

// document is already defined as the head of the xml file
nodes = for (node <- document \\ "version"; if (node.text.contains("SNAPSHOT"))) yeild node

then I want to do something like:

for (node <- nodes) {
    node.text = node.text.split("-")(0)
}

which doesn't work because node is immutable. I looked at the copy method for a Node, but it doesn't include text as a parameter.

Some good answers here. Also see stackoverflow.com/a/23092226/35274 — Philippe
– Philippe, Commented Aug 4, 2016 at 21:08

Daniel C. Sobral · Accepted Answer · 2010-10-25 22:07:47Z

12

You really should take a look at other questions on Stack Overflow about modifying XML. Look at the "Related" links to the right.

Here:

scala> <dependency>
     |     <groupId>foo</groupId>
     |     <artifactId>bar</artifactId>
     |     <version>1.0-SNAPSHOT</version>
     | </dependency>
res0: scala.xml.Elem =
<dependency>
    <groupId>foo</groupId>
    <artifactId>bar</artifactId>
    <version>1.0-SNAPSHOT</version>
</dependency>

scala> new scala.xml.transform.RewriteRule {
     |   override def transform(n: Node): Seq[Node] = n match {
     |     case <version>{v}</version> if v.text contains "SNAPSHOT" => <version>{v.text.split("-")(0)}</version>
     |     case elem: Elem => elem copy (child = elem.child flatMap (this transform))
     |     case other => other
     |   }
     | } transform res0
res9: Seq[scala.xml.Node] =
<dependency>
    <groupId>foo</groupId>
    <artifactId>bar</artifactId>
    <version>1.0</version>
</dependency>

answered Oct 25, 2010 at 22:07

Daniel C. Sobral

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2 Comments

Themerius Over a year ago

Do you have a glue what complexity in time your solution has? How costly is the match/search in the XML-tree? O(log n)?

Daniel C. Sobral Over a year ago

@Themerius It used to be n squared, for depth n, if I recall correctly. There was a ticket and a patch for it, but I don't recall if it ever got fixed.

user unknown · Accepted Answer · 2013-04-29 13:32:05Z

2

Text is represented as a Node inside of the Elements Node. So a bit of functional recursion will let you do a deep copy & filter:

def deepCopy(node:Node) : Node = node match {
  case e : Elem => e.copy(child = this.child.toSeq.map(deepCopy))
  case t : Text => new Text(t.text.split("-").head)
  case x => x
}

disclaimer: this code was not tested for errors

edited Apr 29, 2013 at 13:32

user unknown

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answered Oct 25, 2010 at 20:14

Kevin Wright

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Comments

Erik Kaplun · Accepted Answer · 2014-07-01 16:33:07Z

0

Using Scalate's Scuery CSS3 transforms and scala.xml.Elem#copy:

val xml =
  <dependency>
    <version>1.0-SNAPSHOT</version>
    <version>2.0</version>
  </dependency>

new Transformer {
  $("dependency > version") { node =>
    node.asInstanceOf[Elem].copy(child = Text(node.text.stripSuffix("-SNAPSHOT")))
  }
}.apply(xml)

yields

NodeSeq(<dependency><version>1.0</version><version>2.0</version></dependency>)

answered Jul 1, 2014 at 16:33

Erik Kaplun

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