Here is a snipet from my xsp code:
String xml = appl.getNavigationXML(uriVendorCode);
System.out.println("*********************************** xml is: " + xml);
if (xml != null) {
<util:include-expr><util:expr><xsp:expr>xml</xsp:expr></util:expr></util:include-expr>
}
This includes the xml for navigation within the xml generated by the xsp logic.
In above, the applgetNavigationXML() simply returns a Java string of well formed xml.
Hope this helps. -Yatin
[EMAIL PROTECTED] wrote:
Hi, I'd like to ask, how to insert XML-valid fragment into generated XML document by XSP. I have stored XML fragment (or XHTML or whatever) in the database and I'd like to put it into any element in the serverpages. I tried <element><esql:get-string column="xml_part"/></element> or use ESQL helper or tried to wrote own replacing method, but always is returned text with replaced characters (< to <, > to > etc.).
Another possibility could be maybe use any features of XSL transformation (maybe it could be still better), but I didn't found anything.
Or another (similar) question - how to pull out HTML document (it is not XML valid) from database and consequently send it correctly to the browser?
Do you have any idea? Thanks a lot!
otmar
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-- Yatin Shah, President mailto:[EMAIL PROTECTED] Kripa Inc. http://www.kripa.com Dayton, New Jersey USA phone: 732.329.8303 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Developers of real time event driven distributed DB applications
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