[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] xslt sort remove duplicates
I have an xml file that has a lot of entries which have a name, category and date. I need to display the most recent name of each categoryI To be more precise:.. I did this so far: <xsl:for-each select="exsl:node-set($mobile)/rootb/level1"> <xsl:sort order="descending" select="categ"/> <xsl:sort order="descending" select="date"/> <xsl:value-of select="titlu"/> -<xsl:value-of select="categ"/> - <xsl:value-of select="date"/> </xsl:for-each> right now my list looks like this: it's sorted after categoory, then it's sorted after date. name1- category1 - most recent date * name2 - category1 - date name3 - category1 - date name4 - category1 - date name5 - category1 - date name6 - category2 - most recent date * name7 - category2 - date name8 - category2 - date name9 - category2 - date nameX - category2 - date name6 - category3 - most recent date * name7 - category3 - date name8 - category3 - date name9 - category3 - date nameX - category3 - date ........ I need to dsiplay only the one with the star. So I should only display: name1- category1 - most recent date * name6 - category2 - most recent date * I tried using an if not ((preceding:...but it doesn't work. I think I'm not writing the sintax correctly. Can anyone give some pointers ? 10x name6 - category3 - most recent date * ____________________________________________________________________________________ Moody friends. Drama queens. Your life? Nope! - their life, your story. Play Sims Stories at Yahoo! Games. http://sims.yahoo.com/
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|