Xmlparser error when xml ref's dtd

Question

When calling xmlparser from PL/SQL to parse an xml file
referencing a dtd, I get the following error, even though both
the xml file and dtd are valid with XML Spy:
ERROR at line 1:
ORA-20100: Error occurred while parsing: Invalid argument
ORA-06512: at "SYS.XMLPARSER", line 22
ORA-06512: at "SYS.XMLPARSER", line 69
ORA-06512: at line 6
If I remove the dtd reference from the xml file it runs without
error.
Here's the PL/SQL:
DECLARE
p xmlparser.parser;
d xmldom.DOMDocument;
BEGIN
p := xmlparser.newParser;
xmlparser.parse (p, 'C:\test.xml');
xmlparser.freeParser (p);
END;
Here's the xml file 'C:\test.xml':
<?xml version="1.0" ?>
<!DOCTYPE Department SYSTEM "C:\test.dtd">
<Department>
     <DepartmentRec>
          <SOURCE_KEY>200600 Sales-Northeast</SOURCE_KEY>
          <DEPTID>200600</DEPTID>
          <DEPARTMENT_NAME>SALES-
NORTHEAST</DEPARTMENT_NAME>
     </DepartmentRec>
</Department>
Here's the dtd 'C:\test.dtd':
<!ELEMENT Department (DepartmentRec*)>
<!ELEMENT DepartmentRec (SOURCE_KEY,DEPTID,DEPARTMENT_NAME)>
<!ELEMENT SOURCE_KEY (#PCDATA)>
<!ELEMENT DEPTID (#PCDATA)>
<!ELEMENT DEPARTMENT_NAME (#PCDATA)>
Any help is appreciated!
Thanks - Dana

Answer 1

This error is caused by your using an illegally formatted URL in
the DTD SYSTEM parameter. This is a misleading "feature" of XML
Spy that it allows your file to validate. A line like:
<!DOCTYPE Department SYSTEM "C:\test.dtd">
Is not a legal SYSTEM identifier, which must
be a valid URL. This is a windows-specific file path.
When I changed:
<!DOCTYPE Department SYSTEM "C:\test.dtd">
to
<!-- Use relative URL for DTD -->
<!DOCTYPE Department SYSTEM "test.dtd">
or alternatively to:
<!-- Use absolute URL for DTD -->
<!DOCTYPE Department SYSTEM "file:///C:/test.dtd">
and I changed your file of PL/SQL from:
xmlparser.parse (p, 'C:\test.xml');
to
xmlparser.parse (p, 'file:///C:/test.xml');
it works fine for me.