RE: [XSLTalk] Re: Problem with transforming XML to XM.L using XSL T

OK, I'm not a very frequent user of XMLSpy and I do not have the XSLT
plugin, but I guess you should be able to change the output appearance in
the preferences. If you cannot solve the issue, I highly recommend you to
contact their support. The other reccomendation is to use a trial of
XSelerator - it is a great tool and does not cost a fortune.
 
Good luck! :-)
 
Cheers, Pieter

-----Original Message-----
From: talktao2003 [mailto:talktao2003@yahoo.com]
Sent: Tuesday, July 22, 2003 8:56 AM
To: XSLTalk@yahoogroups.com
Subject: [XSLTalk] Re: Problem with transforming XML to XM.L using XSL T


Hi, 
I guess you are right. I use XMLSPY v5 rel.4 and I got the result by 
pressing F10 or clicking the "XSL Transformation" on the "XSL" menu. 
I also use the IE6.0 to open the XML file directly and I got the same 
result. I spent a whole day on it and couldn't figure it out. What's 
wrong? 

Thanks a lot.


--- In XSLTalk@yahoogroups.com, Pieter Reint Siegers Kort 
<pieter.siegers@e...> wrote:
> Hi again,
>  
> I still do not know how you obtain your output. 
>  
> I mean, if I copy and paste both of your XML and XSL into my copy of
> XSelerator 2.5 and press F9 (Transform) this is what I see in the 
text
> output mode window:
>  
> <catalog>
> <cd>
> <name>Name1</name>
> <artist>Bonnie Tyler</artist>
> </cd>
> <cd>
> <name>Name2</name>
> <artist>Jim Henry</artist>
> </cd>
> </catalog>
>  
> which is exactly what you wish to see - all of your tags are there!
>  
> That's why I asked - how do you OBTAIN your output? Tell us more 
about the
> way you work and what software you use, that'll help clarify the 
situation.
> Something must be filtering your tags out there before you see the
> results...
>  
> Regards, Pieter
> 
> -----Original Message-----
> From: talktao2003 [mailto:talktao2003@y...]
> Sent: Monday, July 21, 2003 4:56 PM
> To: XSLTalk@yahoogroups.com
> Subject: [XSLTalk] Re: Problem with transforming XML to XM.L using 
XSLT
> 
> 
> Hi there,
> Thanks for reply. I mean I cannot see the tags in the XML document 
> generated. The output I got is like this:
> 
> Name1 Bonnie Tyler Name2 Jim Henry 
> 
> As you see, I just got the contents of the each element. where are 
> the element names? like, <cd><name></name></cd>.
> 
> Thanks again.
> 
> 
> --- In XSLTalk@yahoogroups.com, Pieter Reint Siegers Kort 
> <pieter.siegers@e...> wrote:
> > Hi,
> >  
> > Can you tell us how do you exaclty obtain your output? Because I 
> can clearly
> > see that the desired output is exactly what you get when you 
> perform the
> > transformation of your sources.... ¿so what's up?
> >  
> > Regards, Pieter
> > 
> > -----Original Message-----
> > From: talktao2003 [mailto:talktao2003@y...]
> > Sent: Monday, July 21, 2003 12:34 PM
> > To: XSLTalk@yahoogroups.com
> > Subject: [XSLTalk] Problem with transforming XML to XM.L using 
XSLT
> > 
> > 
> > Can anybody let me know why I cannot get the desired output? 
> > Thanks so much.
> > 
> > INPUT XML FILE:
> > 
> > <?xml version="1.0" encoding="UTF-8"?>
> > <?xml-stylesheet type="text/xsl" href="cdcatalog3.xsl"?>
> > <catalog>
> >   <cd>
> >     <name>Name1</name>
> >     <artist>Bonnie Tyler</artist>
> >     <company>CBS Records</company>
> >     <price>9.90</price>
> >     <year>1988</year>
> >   </cd>
> >   <cd>
> >     <name>Name2</name>
> >     <artist>Jim Henry</artist>
> >     <company>London</company>
> >     <price>7.80</price>
> >     <year>1987</year>
> >   </cd>
> > </catalog>
> > 
> > INPUT XSLT FILE:
> > <?xml version="1.0" encoding="UTF-8"?>
> > <xsl:stylesheet version="1.0" 
> > xmlns:xsl=" http://www.w3.org/1999/XSL/Transform
<http://www.w3.org/1999/XSL/Transform> 
> < http://www.w3.org/1999/XSL/Transform
<http://www.w3.org/1999/XSL/Transform> > 
> > < http://www.w3.org/1999/XSL/Transform
<http://www.w3.org/1999/XSL/Transform> 
> < http://www.w3.org/1999/XSL/Transform
<http://www.w3.org/1999/XSL/Transform> > > ">
> > <xsl:output method="xml" omit-xml-declaration="yes" indent="yes"/>
> > <xsl:template match="/">
> > <catalog>
> >   <xsl:for-each select="catalog/cd">
> >     <cd>
> >       <name>    <xsl:value-of select="name"/>  </name>
> >       <artist> <xsl:value-of select="artist"/>  </artist>
> >     </cd>
> >   </xsl:for-each>
> > </catalog>
> > </xsl:template>
> > </xsl:stylesheet>
> > 
> > DESIRED OUPUT:
> > <catalog>
> >   <cd>
> >     <name>Name1</name>
> >     <artist>Bonnie Tyler</artist>
> >   </cd>
> >   <cd>
> >     <name>Name2</name>
> >     <artist>Jim Henry</artist>
> >   </cd>
> > </catalog>
> > 
> > OUTPUT GOT USING THE METHOD ABOVE:
> > Name1 Bonnie Tyler Name2 Jim Henry 
> > 
> > 
> > 
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Received on Tuesday, 22 July 2003 11:15:41 UTC