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

["Dimitre Novatchev dnovatchev@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

On Thu, Jul 7, 2022 at 6:57 PM Dimitre Novatchev dnovatchev@xxxxxxxxx <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> Hi Roger,
>
> > Well, I devised a solution, but it sure ain't simple.
> >
> >   string-join(for $i in reverse(sort(xs:restriction/xs:enumeration, (),
> function($enum) {string-length($enum/@value )})) return $i/@value, '|')
> >
> >   Is there is a simpler XPath expression to solve this problem?
>
> I think this can be written in a good and readable way as below:
>
> string-join(
>                   sort(/*/xs:restriction/xs:enumeration/@value,
>                          (),
>                           function($s) {- string-length($s)}) ,
>                    '|'
>                  )
>


The above produces:
10|11|12|1|2|3|4|5|6|7|8|9


In case it is necessary the values to be in reverse order, I would use
what I think is probably the best, simplest and most direct solution (it's
a pity there is no default prefix for the "math" namespace):

string-join(
                  sort(/*/xs:restriction/xs:enumeration/@value,
                         (),
                          function($s) {- Q{
http://www.w3.org/2005/xpath-functions/math}log10($s)}) ,
                   '|'
                 )

This produces exactly the wanted result. Note: no reverse(), no for/return :

12|11|10|9|8|7|6|5|4|3|2|1


> Thanks,
> Dimitre
>
> On Thu, Jul 7, 2022 at 10:02 AM Roger L Costello costello@xxxxxxxxx <
> xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
>> Hi Folks,
>>
>> I have an XSD simpleType with enumeration values:
>>
>> <xsd:simpleType name="RunwayLightingType">
>>   <xsd:restriction base="xsd:string">
>>    <xsd:enumeration value="1"/>
>>    <xsd:enumeration value="2"/>
>>    <xsd:enumeration value="3"/>
>>    <xsd:enumeration value="4"/>
>>    <xsd:enumeration value="5"/>
>>    <xsd:enumeration value="6"/>
>>    <xsd:enumeration value="7"/>
>>    <xsd:enumeration value="8"/>
>>    <xsd:enumeration value="9"/>
>>    <xsd:enumeration value="10"/>
>>    <xsd:enumeration value="11"/>
>>    <xsd:enumeration value="12"/>
>>   </xsd:restriction>
>>  </xsd:simpleType>
>>
>> I want to turn the values into a regex such that the longest value (not
>> the biggest value, the string longest value) is listed first:
>>
>> 12|11|10|9|8|7|6|5|4|3|2|1
>>
>> Seems like a simple problem, right? Should have a simple solution, right?
>>
>> Well, I devised a solution, but it sure ain't simple.
>>
>> string-join(for $i in reverse(sort(xs:restriction/xs:enumeration, (),
>> function($enum) {string-length($enum/@value )})) return $i/@value, '|')
>>
>> Is there is a simpler XPath expression to solve this problem?
>>
>> /Roger
>>
>>
>>
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-- 
Cheers,
Dimitre Novatchev
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