Subject: [xsl] following-sibling problem From: Andy Carr1 <CARRA@xxxxxxxxxx> Date: Wed, 6 Jun 2007 13:14:24 +0100 |
I have some source xml that looks like this: <References> <Reference>Something</Reference> <Reference>Something Else</Reference> <ReferenceReasonText>Reason for Something Else</ReferenceReasonText> <Reference>Another reference</Reference> <References> I'm applying a template match on References <xsl:apply-templates select="References" /> and <xsl:template match="References/Reference"> <!-- I've set up a test variable to determine if the following-sibling is a <ReferenceReasonText> --> <xsl:variable name="test"> <xsl:choose> <xsl:when test="following-sibling::ReferenceReasonText[1]"> <xsl:text>True</xsl:text> </xsl:when> <xsl:otherwise> <xsl:text>False</xsl:text> </xsl:otherwise> </xsl:choose> </xsl:variable> </xsl:template> The value of the test variable returns the following after transformation: <References> <Reference>Something</Reference> VARIABLE = TRUE <Reference>Something Else</Reference> VARIABLE = TRUE <ReferenceReasonText>Reason for Something Else</ReferenceReasonText> <Reference>Another reference</Reference> VARIABLE = FALSE <References> For the first instance of Reference I would expect the variable to be False, what am I doing wrong ... anyone? I've used following-sibling::xxxxx[1] before and got the results I expected. The only difference that I can see is that this XSLT (v2.0) is wrapped up in some FO using XEP 4.3 to produce PDF. Regards Andy Andy Carr Senior IT Specialist Tel: Internal - 298037 External - 01252 558037 Mail Point M1C IBM Application Services Meudon House, Meudon Avenue, Farnborough, GU14 7NB (Notes) Andy Carr1/UK/IBM@IBMGB (Internet)CARRA@xxxxxxxxxx Unless stated otherwise above: IBM United Kingdom Limited - Registered in England and Wales with number 741598. Registered office: PO Box 41, North Harbour, Portsmouth, Hampshire PO6 3AU
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