You can use something like this:
...
<map:transform src="cocoon:/{id}/{lang}.xsl" />
...
and a pipeline:
<map:match pattern="*/*.xsl">
<!-- Retrieve the source of your XSLT using a generator -->
<map:serialize type="xml"/>
</map:match>
You can also implement cache validity policy in your generator to check if
the XSLT has changed since last request and if not then use the cached
version without need to retrieve it from the DB.
Konstantin
From: "Hubert Holtz" <[EMAIL PROTECTED]>
Hello,
I don't know know how to open new XML files with a certain XSL file with
the help of XSP and my Mysql database in Cocoon.
I have a table called sites with an column for id,lang,XML,XSL, in my xsp-
file I have a string-array called parameters where the 2 parameters of
the typed url (e.g. http://blablabla.com/main.xsp?id=220&lang=eng ) are
saved.
Depending on these two parameters I want to process the certain XML file
with the XSL file which are in my database(as a link of course).
Any idea how to manage this?
I can output the values of the parameters, that's not the problem, but how
do I say Cocoon that it has to process these two files from my database?
Any suggestions or examples or links are welcome.
Thanks.
Homer30
---------------------------------------------------------------------
Please check that your question has not already been answered in the
FAQ before posting. <http://xml.apache.org/cocoon/faq/index.html>
To unsubscribe, e-mail: <[EMAIL PROTECTED]>
For additional commands, e-mail: <[EMAIL PROTECTED]>
---------------------------------------------------------------------
Please check that your question has not already been answered in the
FAQ before posting. <http://xml.apache.org/cocoon/faq/index.html>
To unsubscribe, e-mail: <[EMAIL PROTECTED]>
For additional commands, e-mail: <[EMAIL PROTECTED]>
