[MarkLogic Dev General] xpath string construction

Eric Palmitesta eric.palmitesta at utoronto.ca
Tue Oct 14 08:35:01 PDT 2008

Hi Robert, thanks for the reply.

This won't work considering there may be duplicate names.  I'm not 
working with names at all, it was just an example, but I'm looking into 
a way to uniquely identify portions of my data to make deletions and 
other modification behavior a bit easier.

xdmp:request() and xdmp:request-timestamp() look interesting, but I have 
yet to implement something using these.  Also, seems I might be able to 
use xdmp:lock-acquire() and xdmp:lock-release() on a special document 
which contained an integer I could increment, and abstract it all into a 
nextUniqueId() function, for example.

More research required, though.

Thanks again,


Robert Koberg wrote:
> Hi,
> How about something like this (inline below):
>>>> for $person in doc('/people.xml')/people/person
>>>> return
>>>>   <div>
>>>>     $person/@name
>>>>     <a href="delete.xqy?path={ xdmp:path($person) }>delete</a>
>>>>   </div>
> <a href="delete.xqy?name={ $person/@name }>delete</a>
> then (below)
>>>> This will give me nice delete links like 
>>>> "delete.xqy?path=/people/person[1]", but in the supposed delete.xqy, 
>>>> I'd want to do something similar to:
>>>> let $file := '/people.xml'
>>>> let $person := xdmp:get-request-field('path')
>>>> return
>>>>   xdmp:node-delete(doc($file)/$person)
> let $person-name := xdmp:get-request-field('name')
> return
>   xdmp:node-delete(doc($file)/people/person[@name=$person-name])
> I realize you have simplified things (nicely) for your post, but I would 
> use XQuery to build JavaScript here to put the information you need per 
> person into a JSON object (I realize some requirement might preclude you 
> from using JS). Then add a click event handler for each person link. 
> Then, onclick, use the event handler to find the person's json object 
> and construct an ajax POST or DELETE (which can be converted to XML on 
> the client if needed) to send to the XQY.
> best,
> -Rob
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