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Tricky issue with using xslt with badly formed html...

Answer 1

A:

You could first check that a  has a finishing tag . If it doesn't then you take all text you find until you reach a new tag, that is a , <div>, <li> or anything like that and simply copy it to your xml file where you have addad a full  strukture.

This is how I would do it, might not be the best way but it will work.

Marthin 2010-05-21 22:42:26

Answer 2

+1 A:

This transformation:

<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:output omit-xml-declaration="yes" indent="yes"/>
    <xsl:strip-space elements="*"/>

 <xsl:template match="node()|@*">
     <xsl:copy>
       <xsl:apply-templates select="node()|@*"/>
     </xsl:copy>
 </xsl:template>

 <xsl:template match=
  "div[descendant::div or descendant::p]
  |
   p[descendant::div or descendant::p]
  ">
   <xsl:apply-templates/>
 </xsl:template>

 <xsl:template match=
  "div[descendant::div or descendant::p]/text()
  |
   p[descendant::div or descendant::p]/text()
  ">
   <xsl:element name="{name(..)}"
        namespace="{namespace-uri(..)}">
     <xsl:copy-of select="."/>
   </xsl:element>
 </xsl:template>
</xsl:stylesheet>

when applied on the provided XML document, produces the wanted, correct output:

<html>
   <body>
      <p> text 1 </p>
      <p> text 2</p>
      <p> Here is a list

      </p>
      <ul>
         <ol>
            <li> ListItem1 </li>
            <li> ListItem1 </li>
         </ol>
         <dl>
            <li> dl item </li>
            <li> dl item2 </li>
         </dl>
      </ul>
      <p> I was here</p>
   </body>
</html>

Dimitre Novatchev 2010-05-22 01:02:33

Thanks very much for your help! It is extremely difficult to find people who know xslt. Cheers

yunje 2010-05-25 16:19:25

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Tricky issue with using xslt with badly formed html...

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