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[XQuery Talk Mailing List Archive Home] [By Date] [By Thread] [By Subject] [By Author] [Recent Entries] [Reply To This Message] replacing a node in in-memory XMLJohn Snelson john.snelson at oracle.comTue Nov 6 15:29:29 PST 2007
Hi Robert, You can use XQuery Update to do this really easily: copy $cpy := doc("mydoc") modify replace node $cpy/communitygroup/group/services with <services> <service value="false">1</service> <service value="true">2</service> <service value="false">3</service> </services> return $cpy or copy $cpy := doc("mydoc") modify insert nodes ( <service value="false">1</service>, <service value="false">3</service> ) into $cpy/communitygroup/group/services return $cpy You can leave out the "copy ... modify ... return" (transform expression) if you want the modification done to the document in-place. John Robert Walpole wrote: > Hi, > > I am trying to figure out the best way to replace a node within an > in-memory XML fragment. > > For example, lets say I have the following fragment in memory… > > <communitygroup> > <group> > <name>Test</name> > <services> > <service value="true">2</service> > </services> > </group> > </communitygroup> > > …and want to replace the services node with the following… > > <services> > <service value="false">1</service> > <service value="true">2</service> > <service value="false">3</service> > </services> > > Is there a way of doing this in XQuery? > > Obviously I could use an XSL transformation, which would give me the > result I want, but maybe there is a more efficient way using XQuery? > > Thanks > Rob Walpole > Devon Portal Developer > Email http://x-query.com/mailman/listinfo/talk > Web _http://www.devonline.gov.uk_ > > > ------------------------------------------------------------------------ > > _______________________________________________ > http://x-query.com/mailman/listinfo/talk > http://x-query.com/mailman/listinfo/talk -- John Snelson, Oracle Corporation Berkeley DB XML: http://www.oracle.com/database/berkeley-db/xml XQilla: http://xqilla.sourceforge.net
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