Re: [xsl] building a tabular format of data

Sorry, but I don't understand exactly. Do you have already a template for
<reportdetailinfo>? If not, add <xsl:apply-templates
select="reportdetailinfo"/> to your template matching root node '/' at the
place you want this report. If yes, use the mode parameter for
<apply-templates> and <template>:

<xsl:apply-templates select="reportdetailinfo" mode="report"/>

<xsl:template match="reportdetailinfo" mode="report">
    <!-- .... -->
</xsl:template>

For more help I need your current stylesheet.

Regards,

Joerg

----- Original Message -----
From: "Seema R" <seema.r@xxxxxxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Wednesday, February 13, 2002 6:44 AM
Subject: RE: [xsl] building a tabular format of data


> Thanks Joerg,
>
> It works just great by itself. However, in the stylesheet that I am using
I
> have already set
> <xsl:template match="/"> since I have to get values of other nodes before
I
> get to the reportdetailinfo tag.
> I cannot embed the another <xsl:template> ie; <xsl:template
> match="reportdetailinfo"> within this.
>
> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> version="1.0">
> <xsl:template match="/">
> <html>
> <body>
> <form>
> ...Display some information on patient...
> ... If the report type is textual display accordingly...
> Else
> ... the code as given by you to display tabular format of reports
> </form>
> </body>
> </html>
> </xsl:template>
> </xsl:stylesheet>
>
> Please let me know if there is some way to solve this.
> Thanks,
> Seema.


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