Thank you. This is exactly what I want.


--- On Tue, 7/29/08, Michael Kay <mike@xxxxxxxxxxxx> wrote:

> From: Michael Kay <mike@xxxxxxxxxxxx>
> Subject: RE:  create xml from incoming xml and load it into a variable
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Date: Tuesday, July 29, 2008, 6:52 AM
> Unfortunately the output of xsl:message is
> implementation-defined. Your XSLT
> processor appears to be flattening it to a string, which I
> don't think is a
> particularly good idea, but it's allowed by the spec.
> (Some processors
> discard xsl:message output completely, which is also
> permitted.)
> 
> You can also view the variable by copying it temporarily to
> the result tree:
> 
> <debug variable="var">
>   <xsl:copy-of select="$var"/>
> </debug>
> 
> Michael Kay
> http://www.saxonica.com/ 
> 
> > -----Original Message-----
> > From: sudheshna iyer [mailto:sudheshnaiyer@xxxxxxxxx] 
> > Sent: 29 July 2008 11:30
> > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > Subject: Re:  create xml from incoming xml and
> load it 
> > into a variable
> > 
> > Thank you for the reply.
> > 
> > If I use 		
> > 	<xsl:message>
> > 		var: <xsl:copy-of select="$var"/>
> > 	</xsl:message>	
> > It only prints the following for the below mentioned
> xml:
> > 
> >
> 112005-08-26aaaaaaaaaa11111222005-08-26bbbbb112005-08-26ccccc1
> > 12005-08-26dddddddddd11111
> > 
> > For the xml:
> > <?xml version="1.0"
> encoding="UTF-8"?>
> > <root>    
> >     <subroot id="11111">
> >         <ccc>11</ccc>
> >         <ddd>2005-08-26</ddd>
> >         <eee>aaaaa</eee>
> >          <eee>aaaaa11111</eee>
> >     </subroot>
> >     <subroot id="11111">
> >         <ccc>22</ccc>
> >         <ddd>2005-08-26</ddd>
> >         <eee>bbbbb</eee>
> >     </subroot>
> >     <subroot id="11111">
> >         <ccc>11</ccc>
> >         <ddd>2005-08-26</ddd>
> >         <eee>ccccc</eee>
> >     </subroot>
> >     <subroot id="11111">
> >         <ccc>11</ccc>
> >         <ddd>2005-08-26</ddd>
> >         <eee>ddddd</eee>
> >         <eee>ddddd11111</eee>        
> >     </subroot>    
> > </root> 
> > 
> > I want to see the complete xml, just like the input
> xml along 
> > with element names.
> > 
> > For eg:
> > <root>
> > 	<ccc>11</ccc>
> > 	<ddd>2005-08-26</ddd>
> > 	<eee>aaaaa</eee>
> > 	<eee>aaaaa11111</eee>
> >          ...
> >          
> > with the hiearchy preserved. I am debugging larger xsl
> file 
> > and I don't know whether result-tree in the memory
> is 
> > constructed right or not.
> > 
> > Thank you.
> > 
> > 
> > --- On Tue, 7/29/08, Florent Georges
> <lists@xxxxxxxxxxxx> wrote:
> > 
> > > From: Florent Georges <lists@xxxxxxxxxxxx>
> > > Subject: Re:  create xml from incoming xml
> and load it into a 
> > > variable
> > > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > > Date: Tuesday, July 29, 2008, 5:57 AM
> > > sudheshna iyer wrote:
> > > 
> > >   Hi
> > > 
> > > > In the below xsl, how do I see the entire
> xml that is
> > > in
> > > > variable "var"?
> > > 
> > >   What do you mean exactly?  If you want to
> "see"
> > > it in the result
> > > tree, use xsl:sequence or xsl:copy-of:
> > > 
> > >     <bla-bla>
> > >        <xsl:copy-of
> select="$var"/>
> > >     </bla-bla>
> > > 
> > >   If you want to "see" it on the
> console (or whatever your 
> > environment 
> > > directs messages to), for debugging purpose, you
> can use
> > > xsl:messages:
> > > 
> > >     <xsl:message select="$var"/>
> > >     <!-- or... -->
> > >     <xsl:message>
> > >        VAR: <xsl:copy-of
> select="$var"/>
> > >     </xsl:message>
> > > 
> > >   Regards,
> > > 
> > > --drkm
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > > 
> > >      
> > > 
> >
> ______________________________________________________________________
> > > _______
> > > 
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