Re: [xsl] How to set an element as the context without using a for-each loop?

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

On 01/05/2024 20:36, Leo Studer leo.studer@xxxxxxxxxxx wrote:
> Then I am curious
>>
>> my suggestion with
>>
>> <xsl:sequence select="$record ! (Customer_or_Area_Code, Cycle_Date,
>> Sequence_Number) ! f:convert(.)b/>
>>
>> will produce the same result (order) while any use of / will sort the
>> elements in document order.
>>
> do the following expressions with / produce the same results? (notice
> I changed the order in the ()
>
> B <xsl:sequence select="$record / (Customer_or_Area_Code, Cycle_Date,
> Sequence_Number) / f:convert(.)b/>
>
> <xsl:sequence select="$record / (Cycle_Date, Customer_or_Area_Code,
> Sequence_Number) / f:convert(.)"/>
>

Yes, the step

 B B  / (Customer_or_Area_Code, Cycle_Date, Sequence_Number)

and the step

 B B  / (Cycle_Date, Customer_or_Area_Code, Sequence_Number)

will output the three elements (if they exist) in the input document order