Subject: Re: [xsl] Re: Using XSL for a "world records" table From: Ryan Heise <rheise@xxxxxxxxxxxxx> Date: Sun, 18 May 2003 16:25:45 +1000 |
On Sun, May 18, 2003 at 07:51:57AM +0200, Dimitre Novatchev wrote: > > - notice in the existing records page that two people are tied for first > > place. In this case, they should both have "1" in the "Place" column. > > My current approach fails: when I sort on the time column and use the > > value of position() for place, it unfairly gives one of them second > > place. Is there another way? > > Yes, it is called grouping. You can learn more about the Muenchian method > for grouping at: > http://www.jenitennison.com/xslt/grouping/index.html Thanks Dimitre. It almost works with grouping, but now the record in 3rd place is listed as 2nd place. That is: Place Time Name 1 11 seconds Person 1 1 11 seconds Person 2 2 12 seconds Person 3 My code looks like this: <xsl:for-each select="record[count(. | key('records-by-time', time)[1]) = 1]"> <xsl:sort data-type="number" select="time"/> <xsl:apply-templates select="key('records-by-time', time)"> <xsl:with-param name="place" select="position()"/> </xsl:apply-templates> </xsl:for-each> As you see, I'm using the value of position() for place. Do you know of any way to get around this? > I use the datetime_lib.xsl library, distributed with Xselerator (you can > just download the trial version). > > A long time ago I wrote an XSLT Calendar application using this library -- > this can be obtained at: > > http://www.topxml.com/code/default.asp?p=3&id=v20020711152545 Thanks, I'll have a look at it! Ryan XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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