Subject: Formatting a List into a Grid From: "Noah Nordrum" <noah@xxxxxxxxxxxxx> Date: Fri, 14 Apr 2000 17:23:24 -0400 |
I know I saw this in the list-archives, but I can't find it now :( I am trying to take a List and format it into a Grid. I found an example on MSDN, but, big shocker, was outside of the spec. Basicly, I want to turn this: <rooms> <room> <name>1</name> </room> <room> <name>2</name> </room> <room> <name>3</name> </room> <room> <name>4</name> </room> <room> <name>5</name> </room> <room> <name>6</name> </room> </rooms> into this <tr> <td>1</td><td>2</td> </tr> <tr> <td>3</td><td>4</td> </tr> <tr> <td>5</td><td>6</td> </tr> only minor detail is that I want to do <xsl:apply-templates select="name" /> instead of <xsl:value-of select="name" />. Not sure if that effects the answer, but thought I would point it out now. I know you have to do something with position() % 2, but I am having problems expressing what I am trying to do in well formed XSL. Thank you, Noah XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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Re: call-template name attribute, Noah Nordrum | Thread | passing parameters to an xsl-file, Martin Kammermeier |
Re: call-template name attribute, Noah Nordrum | Date | simple Problem I guess!!!!, sanjeev ramachandra |
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