Re: [xsl] XPath expression to convert XSD enumerations into aregex, longest value first

["Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Am 08.07.2022 um 12:23 schrieb Roger L Costello costello@xxxxxxxxx:
> > At this point we're stuck because the requirements aren't clear.
> > In the example, 12 appeared in the output before 11, but we
> > don't know why: there was nothing in the requirements
> > statement that said it should.
>
> The only requirement is for the longer strings to come first.
>
> So any of these results would be fine:
>
> 12|11|10|9|8|7|6|5|4|3|2|1
> 11|12|10|9|8|7|6|5|4|3|2|1
> 10|11|12|9|8|7|6|5|4|3|2|1
> 12|11|10|1|2|3|4|5|6|7|8|9
> ...
>
> What is the simplest XPath expression to achieve that result?

As I indicated earlier, if you know => as an operator you might consider
the following simpler:

xs:restriction/xs:enumeration/@value/string()
=> sort((), function($s) { -string-length($s) })
=> string-join('|')

But in the end its syntactic sugar.