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Posted to users@cocoon.apache.org by Kirk Storer <st...@muohio.edu> on 2004/03/04 20:24:42 UTC

Easy way to include xhtml when transforming xml to xhtml

Is there an easy way to include xhtml. For example, my xslt will process 
the specified xml document. And some how include the the xhtml file to 
set up the "framework" of the page with the content from the xml file in 
the "main area".

Thanks,
Kirk

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Re: Easy way to include xhtml when transforming xml to xhtml

Posted by Kirk Storer <st...@muohio.edu>.
Thank you so much. That save me a whole lot of time.

Kirk

Markus Heussen wrote:

>Ah now I understand your concern. Use <xsl:copy-of select="content"/> or
><xsl:copy-of select="content/*"/>.
>
>Markus
>
>  
>
>>-----Original Message-----
>>From: Kirk Storer [mailto:storerka@muohio.edu]
>>Sent: Thursday, March 04, 2004 9:00 PM
>>To: users@cocoon.apache.org
>>Subject: Re: Easy way to include xhtml when transforming xml to xhtml
>>
>>
>>Ok, I will try to explain this better.
>>I am using two files as sources, combined via aggregate in the sitemap.
>>The content of one of the files will be contained in <menu></menu>
>>The content of the other file will be contained in <content></content>
>>Both of these are then inside of <page> </page>
>>
>>I want to process the xml within <menu> into xhtml.
>>I merely want to use what is in the <content> tags as it is, it is
>>already xhtml.
>>
>>
>>this is an extremely simplified version of the xsl file
>>----------------------------------------------
>><xsl:template match="page">
>>    <xsl:for-each select="menu">
>>           do whatever
>>    </xsl:for-each>
>>
>>    <xsl:value-of select="content"/>
>></xsl:template>
>>-----------------------------------------------
>>If i use <xsl:value-of select="content"/> it displays the text, but not
>>all of the markup tags
>>
>>How can it get it to include all of the markup tags that are in the
>>file, such as
>> <p> </p>
>> <ul><li></li></ul>
>>etc.
>>
>>Thanks,
>>Kirk
>>
>>
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>>To unsubscribe, e-mail: users-unsubscribe@cocoon.apache.org
>>For additional commands, e-mail: users-help@cocoon.apache.org
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>>
>>    
>>
>
>
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>  
>


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RE: Easy way to include xhtml when transforming xml to xhtml

Posted by Markus Heussen <he...@asspaxx.de>.
Ah now I understand your concern. Use <xsl:copy-of select="content"/> or
<xsl:copy-of select="content/*"/>.

Markus

> -----Original Message-----
> From: Kirk Storer [mailto:storerka@muohio.edu]
> Sent: Thursday, March 04, 2004 9:00 PM
> To: users@cocoon.apache.org
> Subject: Re: Easy way to include xhtml when transforming xml to xhtml
>
>
> Ok, I will try to explain this better.
> I am using two files as sources, combined via aggregate in the sitemap.
> The content of one of the files will be contained in <menu></menu>
> The content of the other file will be contained in <content></content>
> Both of these are then inside of <page> </page>
>
> I want to process the xml within <menu> into xhtml.
> I merely want to use what is in the <content> tags as it is, it is
> already xhtml.
>
>
> this is an extremely simplified version of the xsl file
> ----------------------------------------------
> <xsl:template match="page">
>     <xsl:for-each select="menu">
>            do whatever
>     </xsl:for-each>
>
>     <xsl:value-of select="content"/>
> </xsl:template>
> -----------------------------------------------
> If i use <xsl:value-of select="content"/> it displays the text, but not
> all of the markup tags
>
> How can it get it to include all of the markup tags that are in the
> file, such as
>  <p> </p>
>  <ul><li></li></ul>
> etc.
>
> Thanks,
> Kirk
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: users-unsubscribe@cocoon.apache.org
> For additional commands, e-mail: users-help@cocoon.apache.org
>
>


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Re: Easy way to include xhtml when transforming xml to xhtml

Posted by Kirk Storer <st...@muohio.edu>.
Ok, I will try to explain this better.
I am using two files as sources, combined via aggregate in the sitemap.
The content of one of the files will be contained in <menu></menu>
The content of the other file will be contained in <content></content>
Both of these are then inside of <page> </page>

I want to process the xml within <menu> into xhtml.
I merely want to use what is in the <content> tags as it is, it is 
already xhtml.


this is an extremely simplified version of the xsl file
----------------------------------------------
<xsl:template match="page">
    <xsl:for-each select="menu">
           do whatever
    </xsl:for-each>

    <xsl:value-of select="content"/>
</xsl:template>
-----------------------------------------------
If i use <xsl:value-of select="content"/> it displays the text, but not 
all of the markup tags

How can it get it to include all of the markup tags that are in the 
file, such as
 <p> </p>
 <ul><li></li></ul>
etc.

Thanks,
Kirk


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RE: Easy way to include xhtml when transforming xml to xhtml

Posted by Markus Heussen <he...@asspaxx.de>.
Well, I don't exactly understand what's your concern. But have a look at the
cinclude transformer:
http://cocoon.apache.org/2.1/userdocs/transformers/cinclude-transformer.html

Markus

> -----Original Message-----
> From: Kirk Storer [mailto:storerka@muohio.edu]
> Sent: Thursday, March 04, 2004 8:25 PM
> To: users@cocoon.apache.org
> Subject: Easy way to include xhtml when transforming xml to xhtml
>
>
> Is there an easy way to include xhtml. For example, my xslt will process
> the specified xml document. And some how include the the xhtml file to
> set up the "framework" of the page with the content from the xml file in
> the "main area".
>
> Thanks,
> Kirk
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: users-unsubscribe@cocoon.apache.org
> For additional commands, e-mail: users-help@cocoon.apache.org
>
>


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