Re: How to do macro substitution in XPath?

Hi again,

Just adding as an almost-gratuitous aside that one of the XSLT 3.0 features I
am enjoying is the binding of functions to variables:

<xsl:variable name="is:LBJ" select="function($book as element(Book)) as
xs:boolean {$book/Title eq 'LBJ Bio' and $book/Author eq 'Caro')"/>

"if /Book/$is:LBJ(.) ..."

Like xsl:function except a little more lightweight and flexible, even, nice
for maintainability.

Not only useful in XSLT but Schematron+XSLT as well.

Regards, Wendell

-----Original Message-----
From: Michael Kay mike@xxxxxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Saturday, December 2, 2023 1:13 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re:  How to do macro substitution in XPath?

>
> Here is the wanted "pure XPath" that preserves the "readability":
>
>       let $LBJ := /Book/Title eq 'Path to Power' and /Book/Author eq 'Robert
A. Caro',
>           $RM  := /Book/Title eq 'Power Broker' and /Book/Author eq 'Robert
A. Caro'
>         return
>           if($LBJ) then 'action 1'
>           else if($RM) then 'action 2'
>           else 'Error'
>

The disadvantage here is that the code doesn't have a high level of
reusability. The code as written can only be used to test a book element that
is the outermost element of the current document. I prefer to use functions:

let $LBJ := function($book as element(Book)) as xs:boolean {$book/Title eq
'....' and $book/Author eq '....') let $RM := function(....)

return
   if ($LBJ(/Book)) then 'action 1'
   else if ($RM(/Book)) then 'action 2'
   ...

But although an XPath expression was requested, the context appears to be
XSLT, and in that case I would use good old template rules:

<xsl:template match="Book[Title eq 'Path to Power' and Author eq 'Robert A.
Caro']">
   action-1
</xsl:template>

<xsl:template match="Book[Title eq 'Power Broker' and Author eq 'Robert A.
Caro']">
   action-2
</xsl:template>


Michael Kay
Saxonica