Subject: Re: [xsl] How to handle xsi prefix in XSLT From: "Thomas B. Passin" <tpassin@xxxxxxxxxxxx> Date: Fri, 30 Nov 2001 10:17:27 -0500 |
Using your original xml sample, and this stylesheet, this version of the transformation works fine using either msxml3 or Saxon. You are going to have to tell us what system and processors you use, and how you are trying to produce the transformation, and how you are trying to use it. For example, if you expect the processor to find the stylesheet from the processing instruction in the xml file, that may not be working for a number of reasons. BTW, your method for producing a "Root" element works fine, but it is a little easier to get the same result like this: <xsl:template match='/'> <Root> <xsl:apply-templates/> </Root> </xsl:template> Tom P [Janusz Dalecki] Please forgive my ignorance, I am new to XSLT, I have fixed my XSLT to the one below which inserts a root element (“Root”) and I am still getting the same error – am I doing something stupid? <?xml version = "1.0" encoding = "UTF-8"?> <xsl:stylesheet xmlns:xsl = "http://www.w3.org/1999/XSL/Transform" xmlns:xsi = "http://www.w3.org/2000/10/XMLSchema-instance" version = "1.0"> <xsl:template match = "/"> <xsl:element name = "Root"> <xsl:apply-templates/> </xsl:element> </xsl:template> <xsl:template match = "Command"> <xsl:element name = "Transaction"> <xsl:apply-templates select = "@*"/> </xsl:element> </xsl:template> <xsl:template match = "Command[@xsi:type]"> <xsl:element name = "hhhhhhh"> </xsl:element> </xsl:template> </xsl:stylesheet> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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