Namespace handling in Groovys XmlSlurper

Question

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>?

score 20 · Accepted Answer · answered Dec 29 '11 at 16:13

20

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.

answered Dec 29 '11 at 16:13

Kai Sternad

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1

Thank you, that explains a few things :) I'm parsing different xmls and I do not know which namespaces they are using; is there any possibility to achieve the output of `println xml.':bar'` apart from parsing the root element for namespaces and declaring them? – codeporn Dec 29 '11 at 17:04
1

I think that XmlSlurper is namespace aware by default, since `new XmlSlurper() -> new XmlSlurper(false, true)`: http://stackoverflow.com/questions/33418826/are-xmlparser-and-xmlslurper-namespace-aware-by-default – jalopaba Nov 03 '15 at 09:59
Ugh. For some reason, this isn't working for me. I'm parsing an RSS document with namespaces declared in the headers. I can access rss.channel.link but rss.channel.'atom:link'.text() and rss.channel.'atom:link'.@href.text() all return blank values. :( – Umopepisdn May 15 '16 at 20:55
I fixed it with .declareNamespace('atom':'http://www.w3.org/2005/Atom'). But argh, why should I have to do this when the header is in the xmlns element in the response? – Umopepisdn May 15 '16 at 21:06

score 4 · Answer 2 · answered Oct 24 '17 at 17:32

I know this was answered a while ago, but here's an alternative for anyone else facing the same issue. The XmlSlurper class has three constructors, a couple of which allow you to specify you want it to be namespace-aware.

public XmlSlurper(boolean validating, boolean namespaceAware)

Declare the slurper by calling new XmlSlurper(false, true). I hope this is useful to others.

Namespace handling in Groovys XmlSlurper

2 Answers2

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