在问题中的代码中添加“魔术”给出:
def xml = """<tag id="root">
| <tag id="foo" other="blah" more="meh">
| <tag id="bar" other="huh"/>
| </tag>
|</tag>""".stripMargin()
def root = new XmlSlurper().parseText(xml)
root.breadthFirst().each { n ->
n.replaceNode {
"${n.@id}"( n.children() )
}
}
println groovy.xml.XmlUtil.serialize(root)
哪个打印:
<?xml version="1.0" encoding="UTF-8"?><root>
<foo>
<bar/>
</foo>
</root>
HOWEVER, this will drop any content in the nodes. To maintain content, we would probably need to use recursion and XmlParser to generate a new doc from the existing one... I'll have a think
More general solution
I think this is more generalised:
import groovy.xml.*
def xml = """<tag id="root">
| <tag id="foo" other="blah" more="meh">
| <tag id="bar" other="huh">
| something
| </tag>
| <tag id="bar" other="huh">
| something else
| </tag>
| <noid>woo</noid>
| </tag>
|</tag>""".stripMargin()
def root = new XmlParser().parseText( xml )
def munge( builder, node ) {
if( node instanceof Node && node.children() ) {
builder."${node.@id ?: node.name()}" {
node.children().each {
munge( builder, it )
}
}
}
else {
if( node instanceof Node ) {
"${node.@id ?: node.name()}"()
}
else {
builder.mkp.yield node
}
}
}
def w = new StringWriter()
def builder = new MarkupBuilder( w )
munge( builder, root )
println XmlUtil.serialize( w.toString() )
And prints:
<?xml version="1.0" encoding="UTF-8"?><root>
<foo>
<bar>something</bar>
<bar>something else</bar>
<noid>woo</noid>
</foo>
</root>
Now passes through nodes with no (or empty) id
attributes