Check XML file against XSD? Via XSLT?

Question

Hello,
we´re using a 6.20 system (XI not available) and want to check XML files against a XML-schema definition file (xsd).
Is there a function in 6.20 available for this? Is it possible to use "CALL TRANSFORMATION" for it? I´ve seen other XSLT parser in the internet which are able to check a source XML against an XSD before transforming it to the result XML.
But is this possible with an 6.20 system?
Thank you for your help,
best regards,
Stefan

Answer 1

Here is the sample XML and the schema.
XML:
<?xml version="1.0" encoding="UTF-8"?>
<book isbn="0836217462">
<title>
  Being a Dog Is a Full-Time Job
</title>
<author>Charles M. Schulz</author>
<character>
  <name>Snoopy</name>
  <friend-of>Peppermint Patty</friend-of>
  <since>1950-10-04</since>
  <qualification>
    extroverted beagle
  </qualification>
</character>
<character>
  <name>Peppermint Patty</name>
  <since>1966-08-22</since>
  <qualification>bold, brash and tomboyish</qualification>
</character>
</book>
XSD:
<?xml version="1.0" encoding="utf-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xs:element name="book">
    <xs:complexType>
      <xs:sequence>
        <xs:element name="title" type="xs:string"/>
        <xs:element name="author" type="xs:string"/>
        <xs:element name="character" minOccurs="0" maxOccurs="unbounded">
          <xs:complexType>
            <xs:sequence>
              <xs:element name="name" type="xs:string"/>
              <xs:element name="friend-of" type="xs:string" minOccurs="0"
                      maxOccurs="unbounded"/>
              <xs:element name="since" type="xs:date"/>
              <xs:element name="qualification" type="xs:string"/>
            </xs:sequence>
          </xs:complexType>
        </xs:element>
      </xs:sequence>
      <xs:attribute name="isbn" type="xs:string"/>
    </xs:complexType>
  </xs:element>
</xs:schema>
- Suman