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我想转换一个我用XmlSlurper. （相同的）XML 标记名称应替换为id属性值；应删除所有其他属性。从这段代码开始： 

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)

// Some magic here.

println groovy.xml.XmlUtil.serialize(root)

我想得到以下信息： 

<root>
  <foo>
    <bar/>
  </foo>
</root>

（我在 XML 上编写测试断言，并希望简化它们的结构。）我已经阅读了使用 XmlSlurper 更新 XML 并四处搜索，但没有发现在保留其子节点的同时与节点交换replaceNode()或交换节点的方法。replaceBody()

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1 回答 1

5

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

于 2013-02-18T15:24:35.553 回答