[xsl] RE:

[Kathryn dot Grant at freight dot fedex dot com]

Jeni,

Thanks so much!  That was exactly the problem.  Thanks, Dimitre, as well!

Someone commented earlier what a great list this is--people are so willing
to help.  I really agree, and thank you all.

Kathryn

-----Original Message-----
From: Jeni Tennison [mailto:jeni@jenitennison.com]
Sent: Monday, August 19, 2002 10:40 AM
To: Kathryn.Grant@freight.fedex.com
Cc: xsl-list@lists.mulberrytech.com
Subject: Re:


Hi Kathryn,

> I really appreciate your answer. Unfortunately, I tried it and it
> didn't work. The browser doesn't give me any kind of error message.
> The HTML in the stylesheet shows up fine, but none of the XML data
> shows up. When I take out the parameter and just type in the
> attribute, e.g.,
>
> <xsl:for-each select="//brpfields/record[@S3G >'0']">
> <xsl:sort data-type="number" select="@S3G"/>
>
> The transformation works correctly.

Looking through the code you sent before, I think that the problem
might be that you're setting the parameter with:

>> <xsl:param name="param1" select="S3G"/>

which sets the parameter to the value of the S3G element child of the
root node, whereas you want:

<xsl:param name="param1" select="'S3G'" />
                                 ^   ^

which sets the parameter to the *string* "S3G". Then you can use what
Dimitre suggested:

> <xsl:for-each select="//brpfields/record[@*[name()=$param1] >'0']">
>   <xsl:sort data-type="number" select="@*[name()=$param1]"/>
>   ..........
> </xsl:for-each>

Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list