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Editorial Team
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Asked: 2026-06-13T08:02:33+00:00

<?xml> <env> <body> <folderxml> <folder id=’1′> <folder id=’2′ name =’document’> <folder id=’3′ name =’Music’>

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<?xml>
<env>
<body>
<folderxml>
<folder id='1'>
 <folder id='2' name ='document'>
   <folder id='3' name ='Music'>
     <folder id= '4' name= 'album' xlink='true'>
     </folder>
   </folder>
</folder>
</folder>
</folderxml>
</body>
</env>

I need the path if the attribute ‘link’ is true. Something like xpath.evaluate(path if(link=true)). The result should be something like “/documents/music/album”

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1 Answer

  1. Editorial Team

    After applying the following XPath expression (based on XPath tutorial):

    xmlXPathEval("//*[@xlink='true']", doc)

    you obtain a link to the node list. In the example it is a single node folder with id 4. Then you need to write a function that will travell all the parents and concatenate their name attributes.

    If you needed only the name of the matching folder, than you could go with

    xmlXPathEval("string(//*[@xlink='true']/@name)", doc)

    and read the stringval member of the result. But that matches only the first node.

    Edit: As for the function to iterate over parents I thought of direct access through xmlNodePtr->parent member and reading attributes with xmlGetProp. But as I see your comment (and also the title of the question) I think that you could make use of relative xpath queries, see What’s the most efficient way to do recursive XPath queries using libxml2?.

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