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Home/ Questions/Q 7015823
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T22:42:11+00:00 2026-05-27T22:42:11+00:00

The situation: def str = <foo xmlns:weird=http://localhost/> <bar>sudo </bar> <weird:bar>make me a sandwich!</weird:bar> </foo>

  • 0

The situation:

def str = """
  <foo xmlns:weird="http://localhost/">
    <bar>sudo </bar>
    <weird:bar>make me a sandwich!</weird:bar>
  </foo>
"""
def xml = new XmlSlurper().parseText(str)
println xml.bar

The output of this snippet is

# sudo make me a sandwich!

It seems like the parser merges the contents of <bar> and <weird:bar>.

Is this behavior desired and if yes, how can I avoid this and select only <bar> or <weird:bar>?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T22:42:12+00:00Added an answer on May 27, 2026 at 10:42 pm

    By default XMLSlurper is not namespace aware. This can be turned on by declaring namespaces with the declareNamespace Method.

    def str = """ 
    <foo xmlns:weird="http://localhost/">
      <bar>sudo </bar>
      <weird:bar>make me a sandwich!</weird:bar>
    </foo>
    """ 
    def xml = new XmlSlurper().parseText(str).declareNamespace('weird':'http://localhost/')
    println xml.bar // without namespace awareness, will print "sudo make me a sandwich!"
    println xml.':bar' // will only print "sudo"
    println xml.'weird:bar' // will only print "make me a sandwich!"
    

    The output is:

    sudo make me a sandwich!
    sudo
    make me a sandwich!
    

    The first println will still not be namespace aware. The second println will only print the tag without namespace. If you qualify element with the prefix shown in the third println you only get the namespaced tag.

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