[xsl] Passing a node related to the current node as a template parameter

Hey list,

I have 2 XML documents which I would like to join.
The first one comes as document('arg://messages') and contains
Messages of such structure:

<Messages senderID="25" receiverID="77">
        <Title>test</Title>
        <Content>this message was received by user 77</Content>
</Messages>

The second document contains Persons:

<Person id="77">
        <UserName>admin</UserName>
</Person>

Depending on the situation (whether I'm transforming sent or received
Messages), Persons are passed either as document('arg://senders') or
document('arg://receivers').

Now, I would like to have one common template for Messages to build a
table, no matter if they were sent or received. So my idea is to pass
the Person who has either sent or received the Message as a parameter
for the Message template:

<xsl:template match="Message">
        <xsl:param name="person"/>
        <tr>
                <td>
                        <xsl:value-of select="position()"/>
                </td>
                <td>
                        <xsl:value-of select="$person/UserName"/>
                </td>
                <td>
                        <xsl:value-of select="Title"/>
                </td>
                <td>
                        <xsl:value-of select="Content"/>
                </td>
        </tr>
</xsl:template>

Now, the question is, how do I apply templates on
document('arg://messages') to get a received message table? I need to
join it with document('arg://receivers') on Message/@receiverID =
Person/@id. I was thinking about something along those lines:

<table>
        <xsl:apply-templates select="document('arg://messages')//Message">
                <xsl:with-param name="person"
select="document('arg://receivers')//Person[(_at_)id =
current()/@senderPersonID]"/>
        </xsl:apply-templates>
</table>

But the current() function does not help here because it returns not a
Person but a completely different node (and it should). So how do I
solve it? Or is there a more XSLT-like way to do this?

Thanks,


Martynas
xml.lt

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