sgml to xml conversion

atif picture atif · Dec 15, 2010 · Viewed 9.1k times · Source

I have a following sample sgml data from my .sgm file and I want convert this in to xml

<?dtd name="viewed">
<?XMLDOC>
<viewed >xyz
<cite>
<yr>2010
<pno cite="2010 abc 1188">10
<?/XMLDOC>

<?XMLDOC>
<viewed>abc.
<cite>
<yr>2010
<pno cite="2010 xyz 5133">9
<?/XMLDOC>

Output should be like this:

<index1>
    <num viewed="xyz"/>
    <heading>xyz</heading>
    <index-refs>
      <link  caseno="2010 abc 1188</link>
    </index-refs>
  </index-1>
<index1>
    <num viewed="abc"/>
    <heading>abc</heading>
    <index-refs>
      <link  caseno="2010 xyz 5133</link>
    </index-refs>
  </index-1>

Can this be done in c# or can we use xslt 2.0 to do this kind of conversion?

Answer

Jukka Matilainen picture Jukka Matilainen · Dec 16, 2010

Others have already given some good advice. Here's one way of putting it all together by first converting the input SGML to well-formed XML and then using XSLT to transform that to the exact format you need.

Converting your SGML to well-formed XML

The osx tool from the OpenSP package suggested by mzjn is a good tool for this. Since your SGML markup omits end tags, you need to have a DTD from which the correct nesting of elements can be determined. If you don't have a DTD, you need to create one. For your example input, it could be as simple as this:

<!ELEMENT toplevel o o (viewed)+>

<!ELEMENT viewed - o (#PCDATA,cite)>
<!ELEMENT cite - o (yr,pno)>
<!ELEMENT yr - o (#PCDATA)>
<!ELEMENT pno - o (#PCDATA)>

<!ATTLIST pno cite CDATA #REQUIRED>

You also need to add a proper doctype declaration to the beginning of your SGML file. Assuming you have your DTD in file viewed.dtd.

<!DOCTYPE toplevel SYSTEM "viewed.dtd" >

With this addition, you should now be able use osx to convert the SGML to XML. (It won't be able to convert the processing instructions which start with a / as those are not allowed in XML, and will emit a warning about them.)

osx input.sgm > input.xml

Transforming the resulting XML to your desired format

For the above case, you could use something like the following XSLT stylesheet:

<xsl:stylesheet version="1.0"
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:template match="VIEWED">
    <index1>
      <num viewed="{normalize-space(text())}"/>
      <heading>
        <xsl:value-of select="normalize-space(text())"/>
      </heading>
      <index-refs>
        <xsl:apply-templates select="CITE"/>
      </index-refs>
    </index1>
  </xsl:template>

  <xsl:template match="CITE">
    <link caseno="{PNO/@CITE}"/>
  </xsl:template>

</xsl:stylesheet>