xsl-list
[Top] [All Lists]

[xsl] Simple newbie Q: literal as row title

2009-03-10 11:57:10
As a newbie I have some XSLT working thanks to this group. I have moved to a
"named template" so that I can call template with different arguments - is OK.
But I cannot figure out how to get a "fixed literal" as title of final summary
row.

*************************fragment
    <!-- for each unique audit name -->
    <xsl:for-each  select="Row[(_at_)AuditName and
                generate-id(.)=generate-id(key('audits', @AuditName))]">
        <xsl:call-template name="one-row">
        <xsl:with-param name="AuditN" select="@AuditName"/>
        <xsl:with-param name="findings" select="key('audits', @AuditName)"/>
        </xsl:call-template>
     </xsl:for-each>
     
     <!-- for all audits -->
     <xsl:call-template name="one-row">
         <xsl:with-param name="AuditN" select="TOTAL"/>
         <xsl:with-param name="findings" select="/dsQueryResponse/Rows/Row"/>
         </xsl:call-template>
*************************


Dick Penny






--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--

<Prev in Thread] Current Thread [Next in Thread>