CODEWITHSUNDEEP
xmlgroovy
Anika
Anika

Reputation: 3

Groovy, XMLParser -- Delete XML element while keeping children

I'm trying to modify an XML file by deleting tags (the b tag in the example below) that have been added incorrectly while preserving all children (nested elements and text) of these elements. I was hoping to do this with XMLParser as this is part of a larger script where the order of elements plays a role at a later stage.

Example:

<?xml version="1.0" encoding="UTF-8"?>
<a>
   <b>
      <c />
      Test 1
   </b>
   <b>
      <c />
      Test 2
   </b>
</a>

What I want is something like:

<?xml version="1.0" encoding="UTF-8"?>
<a>
    <c />
    <c />
    Test 1
    Test 2
</a>

What I tried so far is the following (args[0] being a file containing the above example) aiming to simply attach all the children to the parent node:

import groovy.xml.XmlUtil

def f = new File(args[0])
def parser = new XmlParser()
parser.setFeature("http://apache.org/xml/features/nonvalidating/load-external-dtd", false);
parser.setFeature("http://xml.org/sax/features/namespaces", false)

def log = parser.parse(f)
def bs = log.'**'.findAll { node -> (node instanceof Node) && node.name() == 'b' }

bs.each { b ->
    Node bparent = b.parent()
    def bchildren = b.children() 

    bchildren.each { child ->
        if (child instanceof Node) {
            bparent.append(child)
        }
    }

    b.replaceNode{}
}


def xmlu = new XmlUtil()

def result = XmlUtil.serialize(log)
println result

But using this code, I lose the actual text content of the element as XMLParser returns Node and String objects. So the output looks:

<?xml version="1.0" encoding="UTF-8"?>
<a>
    <c />
    <c />
</a>

Is there a way to also add the String objects to the parent node? Or is there perhaps a much simpler approach that I have overlooked so far?

Upvotes: 0

Views: 2308

Answers (2)

Vampire
Vampire

Reputation: 38639

The following will output

<?xml version='1.0' encoding='UTF-8'?>
<a foo='bar'>
  <c blub='bla' />
      Test 1

  <c />
      Test 2

</a>

import groovy.xml.*

def parser = new XmlParser()
parser.setFeature("http://apache.org/xml/features/nonvalidating/load-external-dtd", false);
parser.setFeature("http://xml.org/sax/features/namespaces", false)

def log = parser.parseText('''<?xml version="1.0" encoding="UTF-8"?>
<a foo="bar">
   <b bii="boo">
      <c blub="bla" />
      Test 1
   </b>
   <b>
      <c />
      Test 2
   </b>
</a>''')

def recurse
recurse = { builder, element ->
    element.children().each { child ->
        switch (child) {
            case { (it instanceof Node) && (it.name() == 'b') }:
                recurse builder, child
                break

            case { it instanceof Node }:
                builder."${child.name()}"(child.attributes()) {
                    recurse builder, child
                }
                break

            default:
                builder.mkp.yield child
                break
        }
    }
}

def builder = new MarkupBuilder(new PrintWriter(System.out))
builder.mkp.xmlDeclaration version: '1.0', encoding: 'UTF-8'
builder."${log.name()}"(log.attributes()) {
    recurse builder, log
}

Upvotes: 0

tim_yates
tim_yates

Reputation: 171084

This seems to work... Clone the nodes and add them to the parent, then remove the b from it's parent:

def xmlText = '''<?xml version="1.0" encoding="UTF-8"?>
<a>
   <b>
      <c />
      Test 1
   </b>
   <b>
      <c />
      Test 2
   </b>
</a>'''

import groovy.xml.*

def xml = new XmlParser().parseText(xmlText)
def bs = xml.'*'.findAll { it.name() == 'b' }

bs.each { b ->
    b.children().collect { it instanceof Node ? it.clone() : it }.each { b.parent().children().add(it) }
    b.parent().remove(b)
}

println XmlUtil.serialize(xml)

Upvotes: 2

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