[xquery-talk] replacing a node in in-memory XML

[Michael Kay]

One final comment: note that this method will drop "unused" namespace
declarations from the source document. Not a problem for everyone,
positively welcome in some cases - but unfortunate if your document contains
namespace-sensitive content such as xsi:type attributes. 
 
Michael Kay
http://www.saxonica.com/

  _____  

From: talk-bounces at x-query.com [mailto:talk-bounces at x-query.com] On Behalf
Of Robert Walpole
Sent: 13 November 2007 10:27
To: Wolfgang Meier
Cc: talk at xquery.com
Subject: RE: [xquery-talk] replacing a node in in-memory XML



Just to complete the thread, I have now got this working exactly as I wanted
using a modification of the code provided by Wolfgang. Here is the complete
function:


declare function local:replaceServices($cg as node()) as node()
{
    typeswitch ($cg)
        case $elem as element(services) return (: if the element is a
services element return the following :)
            <services>
            {
                template:populateTemplateElement($elem, true())
            }
            </services>
        case $elem as element() return (: if the element is anything else
just return it as it is - with child nodes intact :) 
           element { node-name($elem) } {
                $elem/@*, for $child in $elem/node() return
local:replaceServices($child)
            }
        default return $cg
};
 
The $cg node is my in-memory node which contains the services element I want
to replace. The function template:populateTemplateElement already existed so
it was much easier for me to use this method than to go down the XSLT route,
which would've meant starting all over.
 
The template:populateTemplateElement function basically compares the atomic
values that exist in the record with all of the possible values specified in
the schema. It then returns elements containing all of the possible atomic
values from the schema with the addition of value attribute containing an
xs:boolean stating whether or not the group has this value.
 
The true() parameter passed to the function is simply to tell the function
whether to return "Yes"/"No" in the value attrib or "true"/"false". In this
case I want "Yes"/"No".
 
Thanks again for all your suggestions and some very interesting comments.
 
Rob Walpole
Devon Portal Developer
Web: http://www.devonline.gov.uk
Email: robert.walpole at devon.gov.uk

  _____  

From: Wolfgang Meier [mailto:wolfgangmm at gmail.com]
Sent: Tue 06/11/2007 15:31
To: Robert Walpole
Cc: talk at xquery.com
Subject: Re: [xquery-talk] replacing a node in in-memory XML



Hi Robert,

> I am trying to figure out the best way to replace a node within an
in-memory
> XML fragment.

I really like to use the typeswitch statement for things like this:

declare function t:replace($node as node()) as node() {
    typeswitch ($node)
        case $elem as element(services) return
            <services>
                <service value="false">1</service>
                <service value="true">2</service>
                <service value="false">3</service>
            </services>
        case $elem as element() return
           element { node-name($elem) } {
                $elem/@*, for $child in $elem/node() return
t:replace($child)
            }
        default return $node
};

t:replace(doc("test.xml")/*)

Wolfgang


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