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How to get the raw XML content under a node #58

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marcelotto opened this issue Feb 27, 2018 · 1 comment
Open

How to get the raw XML content under a node #58

marcelotto opened this issue Feb 27, 2018 · 1 comment

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@marcelotto
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I want to parse the result of a SPARQL XML result like this:

<sparql xmlns="http://www.w3.org/2005/sparql-results#" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.w3.org/2001/sw/DataAccess/rf1/result2.xsd">
  <head>
    <variable name="blurb"/>
  </head>
  <results>
    <result>
      <binding name="blurb">
        <literal datatype="http://www.w3.org/1999/02/22-rdf-syntax-ns#XMLLiteral">
          <p xmlns="http://www.w3.org/1999/xhtml">My name is <b>alice</b></p>
        </literal>
      </binding>
    </result>
  </results>
</sparql>

and get everything under the literal node as a string, i.e. <p xmlns="http://www.w3.org/1999/xhtml">My name is <b>alice</b></p>.

Is this possible with SweetXML somehow?
@doughsay
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Check out this PR: #45

That adds the feature you're looking for. Go bump it and see about getting it merged. I'm trying to get a hold of the maintainers to get the other PRs merged too.

marcelotto added a commit to rdf-elixir/sparql-ex that referenced this issue Mar 19, 2018
@marcelotto
Support for rdf:XMLLiterals in XML decoder
b04912d 
Thanks @doughsay for the hint on kbrw/sweet_xml#45
in kbrw/sweet_xml#58
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2 participants
@doughsay @marcelotto