Re: [xml] Unable to validate a dtd using libxml

[Vidhya Anand <vidhyanarayanaswamy yahoo com>]

--- Daniel Veillard <veillard redhat com> wrote:
> On Tue, Jun 11, 2002 at 08:32:10AM +0530, Vidhya
> Anand wrote:
> > Hi
> >
> > I am unable to validate any dtd using libxml. Used
> description provided
> > by you in a mailing list to validate dtd.
(http://mail.gnome.org/archives/xml/2001-November/msg00178.html)
> > The dtd file(winfodtd.txt) and related error
> messages(error message.txt)
> > are attached. Can you please point out the problem
> with my input? Any
> > help will be appreciated.
>
>   you're trying to parse the standalone DTD with the
> XML content parser
> no surprize it doesn't work !
>   Use xmlParseDTD().
>
> Daniel
>
> -- 
> Daniel Veillard      | Red Hat Network
> https://rhn.redhat.com/
> veillard redhat com  | libxml GNOME XML XSLT toolkit
>  http://xmlsoft.org/
> http://veillard.com/ | Rpmfind RPM search engine
http://rpmfind.net/

Thanks for your prompt response.

I am using xmlParseDTD as described in your mail. My
code reads as follows and the behaviour is as
described in previous mail. I am unable to figure my
problem:


if ((dtd = xmlParseDTD(NULL, (const xmlChar *)
             dtdvalid.c_str())) == NULL)
{
    printf ("Could not parse DTD\n") ;
}
else
{
    xmlValidCtxt cvp;
    cvp.userData = (void *) stderr;
    cvp.error    = (xmlValidityErrorFunc) fprintf;
    cvp.warning  = (xmlValidityWarningFunc) fprintf;
    if (!xmlValidateDtd(&cvp, _doc, dtd))
    {
        printf ("XML contents does not validate
                  against DTD\n");
    }
}





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