How to get the root element with XPath?

Question

Here is my XML document:

<?xml version="1.0" encoding="UTF-8"?>
<container version="1.0" xmlns="urn:oasis:names:tc:opendocument:xmlns:container">
   <rootfiles>
        <rootfile full-path="content.opf" media-type="application/oebps-package+xml"/>
    </rootfiles>
</container>

All I want to do is get the root file element. I'm using TouchXML on the iPhone and here is my xPath query:

/container/rootfiles/rootfile

It doesn't work! I've tried everything.

Good Question (+1). See my answer for pointers and also for a pure XPath solution. :) — Dimitre Novatchev, May 01 '10 at 01:18

score 2 · Accepted Answer · edited May 23 '17 at 10:27

2

That xmlns bit within the container node is a namespace prefix that you may have your query register.

I don't know TouchXML well enough to say what the solution is, but the relevant libxml2 call is xmlXPathRegisterNs() if you go into the innards of TouchXML.

This SO answer might also help point you to the right TouchXML method to use to register this namespace.

edited May 23 '17 at 10:27

Community

1
1

answered Apr 30 '10 at 23:00

Alex Reynolds

95,983
54
240
345

dude, I think you're right on. Apparently touch xml has some problems that are know with xPath. I'm gonna try to fix them! Who knows, maybe someone will benefit? – Sam Stewart Apr 30 '10 at 23:32
That's a nice thing about open source. Good luck! – Alex Reynolds Apr 30 '10 at 23:41

score 1 · Answer 2 · answered May 01 '10 at 01:16

1

I am not a specialist in TouchXML, but this may help you.

There is also a somewhat ugly pure XPath solution.

Use:

/*[name()='container']/*[name()='rootfiles']/*[name()='rootfile']

answered May 01 '10 at 01:16

Dimitre Novatchev

240,661
26
293
431

How to get the root element with XPath?

2 Answers2