xsl:sequence not doing what I thought it would

mphare · August 19th, 2010, 10:51 AM

I am using Saxon 9 and my 3rd edition of XSLT 2.0 is looking mighty dog-eared :)

I must be interpreting the passages (paraphrased):

Quote:

The <xsl:sequence> element is used to deliver an arbitrary sequence[..]. It is [one of] the only XSLT function(s) [..] that can return references to existing nodes [..].

to mean what I'd like them to mean, rather than what they really mean. That, or I'm just using <xsl:sequence> incorrectly.

I would like to replace:

Code:

<xsl:variable name="myPMTable" select="$repoSubTree//pmtable[./pmtitle/@Id eq $myIDRef]"/>

with:

Code:

<xsl:variable name="myPMTable" select="nmirepo:findPMTable($repoSubTree, $myIDRef)"></xsl:variable>

and use:

Code:

    <xsl:function name="nmirepo:findPMTable">
        <xsl:param name="basenode"></xsl:param>
        <xsl:param name="idref"></xsl:param>
        
        <xsl:sequence select="$basenode//pmtable[./pmtitle/@Id eq idref]"/>
    </xsl:function>

as the function so I can use it is multiple places in the stylesheet, and in other stylesheets.

Seems logical to me, returning a reference to existing nodes. But I don't get anything back from the function. Certainly not a pointer to the table I'm looking for.

So, what am I missing with <xsl:sequence>?

Thanks,

samjudson · August 19th, 2010, 11:03 AM

Might it simply be that you are missing a '$' in front of idref?

Code:

<xsl:sequence select="$basenode//pmtable[./pmtitle/@Id eq $idref]"/>

mphare · August 19th, 2010, 11:07 AM

Well, that's twice in one week.
I think I need a vacation. I'm looking at this stuff cross-eyed.

Thanks!