Re: arbitrary sorting (Part IV)

> but you could presumably use count(preceding-sibling::*)+1

Yeah, that's it, thanks!
This morning on my way to the university I had the same idea ... :-)
Btw, for using the expression as a sort key there's no need to add 1.

Best regards,
Oliver


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|  ob|do        Dipl.Inf. Oliver Becker                             |
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