Creation of a XML Document with DTD entry

Question

I am trying to create a XML file on the fly. I use the
NodeFactory to contruct a document (createDocument()). I have
then setStandalone("no") and setVersion("1.0"). when I try to
add a DOCTYPE node via appendChild( new XMLNode("test",
Node.DOCUMENT_TYPE_NODE)), I get a ClassCastException.
This is using the 1.0.1.1.0 xml parser.
What is the mechanism to add a node of this type? I noticed that
the NodeFactory did not have a machanism for creating a DOCTYPE
node.
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Answer 1

Nestor Dutko (guest) wrote:
: I am trying to create a XML file on the fly. I use the
: NodeFactory to contruct a document (createDocument()). I have
: then setStandalone("no") and setVersion("1.0"). when I try to
: add a DOCTYPE node via appendChild( new XMLNode("test",
: Node.DOCUMENT_TYPE_NODE)), I get a ClassCastException.
: This is using the 1.0.1.1.0 xml parser.
: What is the mechanism to add a node of this type? I noticed
that
: the NodeFactory did not have a machanism for creating a DOCTYPE
: node.
There is no mechanism to create a new DOCUMENT_TYPE_NODE object
via DOM APIs. The only way to get a DTD object is to parse the
DTD file or the XML file using the DOMParser, and then use the
getDocType() method. Note that new XMLNode("test",
Node.DOCUMENT_TYPE_NODE) does not create a DTD
object. It creates an XMLNode object with the type set to
DOCUMENT_TYPE_NODE, which infact should not be allowed. The
ClassCastException is raised because appendChild expects a DTD
object (based on the type).
Also, we do not do any validation while creating the DOM tree
using the DOM APIs. So setting the DTD in the Document will not
help in validating the DOM tree that is constructed. The only way
to validate an XML file is to parse the XML document using
DOMParser or SAXParser.
Oracle XML Team
http://technet.oracle.com
Oracle Technology Network
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