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Subject: [xsl] RE: From: Kathryn.Grant@xxxxxxxxxxxxxxxxx Date: Mon, 19 Aug 2002 10:56:21 -0700 |
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@xxxxxxxxxxxxxxxx] Sent: Monday, August 19, 2002 10:40 AM To: Kathryn.Grant@xxxxxxxxxxxxxxxxx Cc: xsl-list@xxxxxxxxxxxxxxxxxxxxxx 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
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