Hi Huib,
Thanx for the reply.
I’ve seen this kind of solution before on the internet but this does not work
in my case.
I don’t have a single @ref-attribute in $node-element. Instead I have several
ref-child-elements within a $node-element. (that is what a meant with linear
versus iter
Hi Rob,
I think XQuery and XSLT are very much capable of handling your example. If I
understand it correctly, I do not know what linear or iterative recursion is.
In fact I programmed something similar to the algorithm you describe in XQuery
recently. I was processing nodes from the BaseX datab
Thanks Michael. I ended up with something similar :-)
best
*P
On 7/17/12 9:10 AM, Michael Seiferle wrote:
> Hi Pascal,
>
> maybe you already tried that; but:
>
> let $somethings := for $varGrp at $pos in local:getVariables($from,$to)
> group by $name := string($varGrp/@name)
> return
> for
Hi Pascal,
maybe you already tried that; but:
let $somethings := for $varGrp at $pos in local:getVariables($from,$to)
group by $name := string($varGrp/@name)
return
for $something at $pos in $somethings
return...
Hope this helps, this is the best I can think of at the moment :)
Kind rega
Hi,
Can you please help me solving a challenge I have with a specific type
XML-query?
Ive a bunch of XML-files that are referencing each other.
So, a XML-file can reference to multiple XML-files and a XML-file can be
referenced to by multiple XML-files. This whole reference-structure cou
All:
In a group by query expression like
for $varGrp at $pos in local:getVariables($from,$to)
group by $name := string($varGrp/@name)
return
is there any way to get a value or position for the group count?
In my particular use case, I used this to generate an HTML table and I
would like to
Hi,
Not that I tested or tried it, but it seems to me to be a matter
operator priorities, hence of brackets around both db:replace functions:
( db:replace(...) , db:replace(...) ).
Otherwise the XQuery parser will parse it as
updating function ... {
( let ... for ... let ... let ..
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