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RE: [xsl] Transforming XML to HTML Table
Subject: RE: [xsl] Transforming XML to HTML Table From: "Nick Foster" <Nick.Foster@xxxxxxxxxxx> Date: Wed, 29 Jan 2003 14:09:15 -0000 |
Oleg, No it's not a typo, the <sample> nodes may appear in any order within each <row> node so that the order in the first row might be Sample 1, Sample 2, Sample 3 but in the next row it could be Sample 2, Sample 3, Sample 1. If they were always in the same order then there would be no problem in just looping through each and writing them out in the table. In each row I need to select the <sample> nodes in the same order that they appeared in the first row. Thought: Is it possible to do a sort within each row so that they are all in the same order, alphabetical order perhaps? Cheers, Nick -----Original Message----- From: Oleg Tkachenko [mailto:olegt@xxxxxxxxxxxxx] Sent: 29 January 2003 13:00 To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Transforming XML to HTML Table Nick Foster wrote: > but I only got the value of the first sample in all columns. Is there a > way of looping through the $columntitles/labels and selecting the sample > where the label matches in the current row? Try <xsl:for-each select="$columntitles"> <TD> <xsl:value-of select="$samples[label=current()/label]/displayval"/> </TD> </xsl:for-each> But what's order? You've got "sample 1" in first row and "Sample 1" in the second one. Is this a typo? -- Oleg Tkachenko eXperanto team Multiconn Technologies, Israel XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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