Subject: [xsl] Simple newbie Q: literal as row title From: "Dick Penny" <d_penny@xxxxxxx> Date: Tue, 10 Mar 2009 08:56:41 -0700 |
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[@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
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