How do I refer to the value of the child of a current node when selecting another node with XPath? (DDEX Related)

Question

Consider this XML structure, a dumbed-down version of the DDEX standard:

<doc>
<master>
 <ResourceInfo>
  <Name>Foo</Name>
  <Seq>1</Seq>
 </ResourceInfo>
 <ResourceInfo>
  <Name>Bar</Name>
  <Seq>2</Seq>
 </ResourceInfo>
</master>
<track>
 <Resource>
  <Name>Foo</Name>
 </Resource>
</track>
<track>
 <Resource>
  <Name>Bar</Name>
 </Resource>
</track>
</doc>

I'd like to select the ResourceInfo node in <master> with a child <Name> matching the text value of the Name of each of the track node to get the Seq number.

I can do so directly by getting an lxml tree of each track and explicitly requesting <ResourceInfo>'s like this:

track.xpath('/doc/master/ResourceInfo/Seq[../Name[text()="Foo"]]')

But that assumes I know the name of each track and can explicitly state it ahead of time. I'd like to be able to dumbly map this and somehow replace the "Foo" in the xpath with some reference to the Name text() of current track's Resource.

It's kind of like joining tracks and resources on the text() of the Name in master with the text() of Name in each track. Is there an easy way of doing this with XPath?

I'm trying to iterate over each track, and pull the Seq from the track. Therefore, I can't explicitly ask for "Foo". I need to introspect - "Give me the Seq that is a sibling of a <Name> node in master with a value matching <Name> of the current node in <track>".

What's your expected output? What are you really trying to do? Joining with the tracks doesn't seem very helpful in here, probably don't get the point yet. — Jens Erat, Feb 16 '13 at 00:16
Jens, I'm trying to iterate over each track, and pull the Seq from the . Therefore, I can't explicitly ask for "Foo". I need to introspect - "Give me the Seq that is a sibling of a Name node in master with a value that matches the Name of the current node in ". — Gritty Kitty, Feb 16 '13 at 00:38
The expected output is the same (1 and 2, respectively), but programmatically, I don't want to ask for "Foo" or "Bar". I'd rather XPath interpolate "Name of current node" in so that the XPath expression can remain the same for finding the `` for any track. — Gritty Kitty, Feb 16 '13 at 00:44

score 2 · Answer 1 · answered Feb 16 '13 at 03:24

2

I'm not sure if I understand completely, but if the current context is:

/doc/track/Resource/Name

and you use the following XPath:

/doc/master/ResourceInfo[Name = current()]/Seq

you should get the Seq of the ResourceInfo of the same Name.

answered Feb 16 '13 at 03:24

Daniel Haley

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Daniel, I need something that works with Python. That's super simple looking and great, but it doesn't work with lxml or ElementTree. – Gritty Kitty Feb 16 '13 at 16:26

dlink · Answer 2 · 2013-06-28T22:30:44.780

After reading your comment I now understand what you are after. Uou can simply use Python to do the join:

from lxml import etree

doc = etree.parse('sample.xml')

# gather resources
resources = {}
for element in doc.xpath('/doc/master/ResourceInfo'):
    name = element[0].text
    seq  = element[1].text
    resources[name] = seq

# gather tracks
tracks = []
for element in doc.xpath('/doc/track/Resource/Name'):
    name = element.text
    tracks.append(name)

# join:

for track in tracks:
    print 'Track: %s, seq: %s' % (track, resources.get(track))

# returns: 
# Track: Foo, seq: 1
# Track: Bar, seq: 2

Previous Answer:

The XML was not well formed:

<doc>
  <master>
    <ResourceInfo>
      <Name>Foo</Name>
      <Seq>1</Seq>
    </ResourceInfo>
    <ResourceInfo>
      <Name>Bar</Name>
      <Seq>2</Seq>
    </ResourceInfo>
  </master>
  <track>
    <Resource>
      <Name>Foo</Name>
    </Resource>
  </track>  <!-- was missing backslash -->
  <track>
    <Resource>
      <Name>Bar</Name>
    </Resource>
  </track>
</doc>

Your code works:

from lxml import etree

doc = etree.parse('a.xml')

for element in doc.xpath('/doc/master/ResourceInfo/Seq[../Name[text()="Foo"]]'):
    #print etree.tostring(element)
    print element.text  

# returns
# 1

Yes, it does, but I don't want to explicitly state "Foo". I want to replace "Foo" with "The text of the current node's child". — Gritty Kitty, Feb 16 '13 at 00:37

How do I refer to the value of the child of a current node when selecting another node with XPath? (DDEX Related)

2 Answers2