Scala: XML parsing namespaced attributes

Question

In the example below, how do I access the attribute 'id' once it has a namespace prefix?

scala> val ns = <foo id="bar"></foo>    
ns: scala.xml.Elem = <foo id="bar"></foo>

scala> ns \ "@id"                   
res15: scala.xml.NodeSeq = bar

Above works fine. According to the docs below should work but it doesn't.

scala> val ns = <foo xsi:id="bar"></foo>
ns: scala.xml.Elem = <foo xsi:id="bar"></foo>

scala> ns \ "@{xsi}id"                   
res16: scala.xml.NodeSeq = NodeSeq()

All on Scala 2.8.0.final

Cheers

Answer: It seems without an xlmns in the xml you can't access the attribute. So for the above example to work it needs to be inside an xlm namespace. e.g.:

scala> val xml = <parent xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <foo xsi:id="bar"></foo></parent>
xml: scala.xml.Elem = <parent xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <foo xsi:id="bar"></foo></parent>

scala> xml \ "foo" \ "@{http://www.w3.org/2001/XMLSchema-instance}id"
res3: scala.xml.NodeSeq = bar

From the Scala API (NodeSeq): ns \ "@{uri}foo" to get the prefixed attribute "pre:foo" whose prefix "pre" is resolved to the namespace "uri". But ns \ "@{xsi}id" is not working :( — Thomas Rawyler
– Thomas Rawyler, Commented Feb 22, 2011 at 15:12
Thanks Thomas, I somehow missed that in the docs. I've updated the example to show how it still doesn't work even when following the API correctly. — David
– David, Commented Feb 22, 2011 at 15:33

Community · Accepted Answer · 2017-05-23 12:03:58Z

14

Take a look at this post: Accessing XML attributes with namespaces.

It looks like the uri that is referred to in:

ns \ "@{uri}foo"

Refers to the part after the equal sign. This works:

scala> val ns = <foo xmlns:id="bar" id:hi="fooMe"></foo>
ns: scala.xml.Elem = <foo id:hi="fooMe" xmlns:id="bar"></foo>

scala> ns \ "@{bar}hi"
res9: scala.xml.NodeSeq = fooMe

So I think the first thing after foo is to define your URL and namespace, and then to define the attribute, so if you want to get the attribute "bar", maybe something like this:

scala> val ns = <foo xmlns:myNameSpace="id" myNameSpace:id="bar"></foo>
ns: scala.xml.Elem = <foo myNameSpace:id="bar" xmlns:myNameSpace="id"></foo>

scala> ns \ "@{id}id"
res10: scala.xml.NodeSeq = bar

Although I'm not sure about the correctness of reusing the URI as attribute name.

edited May 23, 2017 at 12:03

CommunityBot

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answered Feb 22, 2011 at 16:00

CaffiendFrog

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2 Comments

David Over a year ago

Hi CaffiendFrog. You're right, I have to use the xmlns as part of the pattern. Thank you.

Dr. Hans-Peter Störr Over a year ago

You might want to mention ns \@ "{id}id" giving the String value of the attribute.

Eugeny Belousov · Accepted Answer · 2018-05-08 15:00:26Z

4

You can use

ns.attributes.head.asAttrMap("xsi:id")

answered May 8, 2018 at 15:00

Eugeny Belousov

411 bronze badge

1 Comment

DenisNovac Over a year ago

Thanks! It runs nice if you work with single node and do not have the namespace definition.

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