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Nokogiri/Xpath namespace query

开发者 https://www.devze.com 2023-02-04 21:50 出处:网络
I\'m trying to pull out the dc:title element using an xpath. I can pull out the metadata using the following code.

I'm trying to pull out the dc:title element using an xpath. I can pull out the metadata using the following code.

doc = <<END
<?xml version="1.0" encoding="UTF-8"?>
<package xmlns="http://www.idpf.org/2007/opf" version="2.0">
  <metadata xmlns:dc="URI">
    <dc:title>title text</dc:title>
  </metadata>
</package>
END

doc = Nokogiri::XML(doc)

# Awesome this works!
puts '//xmlns:metadata'
puts doc.xpath('//xmlns:metadata')
# => <metadata xmlns:dc="URI"><dc:title>title text</dc:title></metadata>

As you can see the above appears to work correctly. However I don't seem to be able to get the title information from this node tree, all of the below fail.

puts doc.xpath('//xmlns:metadata/title')
# => nil

puts doc.xpath('//xmlns:metadata/dc:title')
# => ERROR: `evaluate': Undefined na开发者_JAVA百科mespace prefix

puts doc.xpath('//xmlns:dc:title')
# => ERROR: 'evaluate': Invalid expression: //xmlns:dc:title

Could someone please explain how namespaces should be used in an xpath with the above xml doc.


All namespaces need to be registered when parsing. Nokogiri automatically registers namespaces on the root node. Any namespaces that are not on the root node you have to register yourself. This should work:

puts doc.xpath('//dc:title', 'dc' => "URI")

Alternately, you can remove namespaces altogether. Only do this if you are certain there will be no conflicting node names.

doc.remove_namespaces!
puts doc.xpath('//title')


With properly registered prefix opf for 'http://www.idpf.org/2007/opf' namespace URI, and dc for 'URI', you need:

/*/opf:metadata/dc:title

Note: xmlns and xml are reserved prefixes that can't be bound to any other namespace URI than the built-in 'http://www.w3.org/2000/xmlns/' and 'http://www.w3.org/XML/1998/namespace'.


As an alternative to explicitly constructing a hash of namespace URIs, you can retrieve the namespace definitions from the xml element where they're defined.

Using your example:

# First grab the metadata node, because that's where "dc" is defined.
metadata = doc.at_xpath('//xmlns:metadata')

# Pass metadata's namespaces as the resolver.
metadata.at_xpath('dc:title', metadata.namespaces)

Note that the second xpath could've also been:

doc.at_xpath('//dc:title', metadata.namespaces).to_s

But why search from the root when you have a nearer ancestor? Also, you should consider the namespace-defining element plus its children as the "scope" of the namespace. Searching a limited scope is less confusing and avoids subtle bugs.

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