Subject: Re: [xsl] To display the output depending upon condition From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Sat, 20 Oct 2001 15:14:37 +0100 |
Hi Srini, > 2. At this point I need to check whether that > particular parentId !=0 and if yes I need to > display that message as a particular left not a > part of tree. You can tell which messages are being left out by finding those messages where the parentId is (a) not equal to 0 and (b) not the value of the dbMessageId of any of the other Message elements. Presumably you want these messages to be treated like top-level messages that have a parentId of 0. Since there's no Message with an Id of 0 anyway, you can do this in one test - find all Message elements for which there are no Messages with an Id equal to its parentId: Message[not(@parentId = ../Message/@dbMessageId)] I'd change the template for Messages to select these instead of just those with a parentId of 0: <xsl:template match="Messages"> <xsl:apply-templates select="Message[not(@parentId = ../Message/@dbMessageId)]" /> </xsl:template> I hope that helps, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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