Re: [xslt] transformation removes DOCTYPE declaration form original document - Yes it does!

[Mike Trotman <mike trotman datalucid com>]

OK - I tried this with Perl - and the same things happens!

I am sure this cannot be 'correct' behaviour - unless there is a note in 
the documentation I have missed that says XML::LibXSLT will change your 
SOURCE document?

     1 #!/usr/bin/perl
     2 use strict;
     3 use warnings;
     4 use XML::LibXML;
     5 use XML::LibXSLT;
     6
     7 my $doc = '<?xml version="1.0" encoding="UTF-8"?>
     8 <!DOCTYPE xbel PUBLIC "+//IDN python.org//DTD XML Bookmark 
Exchange Language 1.0//EN//XML" 
"http://pyxml.sourceforge.net/topics/dtds/xbel-1.0.dtd";>
     9 <xbel>
    10 </xbel>
    11 ';
    12
    13 my $xsl = '<?xml version="1.0" encoding="ISO-8859-2"?>
    14 <xsl:stylesheet version="1.0"
    15  xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
    16
    17  <xsl:output method="xml" indent="yes" encoding="iso-8859-2"/>
    18
    19  <xsl:template match="/">
    20   <xsl:apply-templates/>
    21  </xsl:template>
    22
    23  <xsl:template match="/xbel">
    24    <z/>
    25  </xsl:template>
    26
    27 </xsl:stylesheet>
    28 ';
    29
    30 my $parser=XML::LibXML->new();
    31 my $XDOC = $parser->parse_string($doc);
    32 my $XDOC_DUP = $parser->parse_string($doc);
    33 $XDOC->setEncoding('UTF8');
    34
    35 print "Source DOC Before Transform:\n\n";
    36 print $XDOC->toString(1);
    37
    38 my $xslt=XML::LibXSLT->new();
    39 my $xxsl = $parser->parse_string($xsl);
    40 $xxsl = $xslt->parse_stylesheet($xxsl);
    41 my $result=$xxsl->transform($XDOC);
    42
    43 print "\nSource DOC After Transform:\n\n";
    44 print $XDOC->toString(1);
    45 print "\n\nWhereas I was expecting:\n\n";
    46 print $XDOC_DUP->toString(1);



[root www Examples]# perl t1.pl
Source DOC Before Transform:

<?xml version="1.0" encoding="UTF8"?>
<!DOCTYPE xbel PUBLIC "+//IDN python.org//DTD XML Bookmark Exchange 
Language 1.0//EN//XML" 
"http://pyxml.sourceforge.net/topics/dtds/xbel-1.0.dtd";>
<xbel>
</xbel>

Source DOC After Transform:

<?xml version="1.0" encoding="UTF8"?>
<xbel>
</xbel>


Whereas I was expecting:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xbel PUBLIC "+//IDN python.org//DTD XML Bookmark Exchange 
Language 1.0//EN//XML" 
"http://pyxml.sourceforge.net/topics/dtds/xbel-1.0.dtd";>
<xbel>
</xbel>
[root www Examples]#

SO it looks like the XSLT process does change the source document!

Glad you pointed that out


Mike Trotman wrote:

> I'm not sure of the python libxslt syntax - but assume it is like the 
> Perl XML::LibXML
>
> $stylesheet->transform($doc, foo => "value);
>
> which 'throws away' the XSLT output.
>
> So - in the example code RadosÅaw doesn't use the output of the XSLT 
> transformation.
> He just prints the INPUT document twice.
> So it looks like the code HAS changed the input document.
> Which surely has nothing to do with any XSLT / XPATH concerns.
>
> I will try this in Perl tomorrow and see if the same things happens.
>
> Mike
>
> Vincent Lefevre wrote:
>
> > On 2005-04-14 11:24:49 +0200, RadosÅaw Kintzi wrote:
> >  
> >
> > > I don't care about DT declaration in target document. I just care
> > > about one in original (source) document. It has such declaration,
> > > which was removed by processor during transformation.
> > > In other words: the processor change source document. It is right 
> > > behavior or it is a bug?
> > >    
> >
> > This doesn't have much sense. An XSLT processor takes a source
> > document and outputs a target document. Do you mean that your
> > source file is modified, in addition to the output of the target
> > file? I've never seen such a thing...
> >
> >  
>
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