Subject: Re: [xsl] XPATH or 2 templates ? From: David Carlisle <davidc@xxxxxxxxx> Date: Mon, 18 Nov 2002 13:19:49 GMT |
> <?xml-stylesheet type="text/xsl" href="E:\Zvxml\XtraTest\Test.xsl"?> If your processor finds the stylesheet from that href it is being unnecessarily kind to you. It should be a URI, the above refers to some unregistered uri scheme E: it should be file:///E:/Zvxml/XtraTest/Test.xsl" > Why can I not have template match on <Concat> and <Concat/*> within the same > stylesheet ? You can, why do you think you can not? > <xsl:text> </xsl:text> You should use <xsl:text> </xsl:text> to get a newline, using 10 will give you an XML normalised line end which will most likely be output by as a line end suitable for your machine type. If you generate a 13 then the results are less predictable, most likely you will just get a single character 13, which may or may not be the right line end marker for your machine type. I _think_ you want one line for each child of <Concat> so your second part should look like <xsl:for-each select="*"> <xsl:text> </xsl:text> <!-- and on this line you want all the descendent text separated by ; --> <xsl:for-each select=".//*[not(*)]"> <xsl:value-of select="."/> <xsl:if test="position()!=last()">;</xsl:if> </xsl:for-each> </xsl:for-each> David _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Scanning Service. For further information visit http://www.star.net.uk/stats.asp or alternatively call Star Internet for details on the Virus Scanning Service. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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