this is the full xsl code:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<!-- overload the text template,
do NOT automatically pass-thru text -->
<xsl:template match="text()">
</xsl:template>
<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>
<!-- for all SEG*/NAME elements, display the name, followed by a * -->
<xsl:template match="*[starts-with(name(), 'SEG')]/NAME">
<xsl:value-of select="."/><xsl:text>*</xsl:text>
</xsl:template>
</xsl:stylesheet>
will that not go thru and apply all the templates that have matches?
-L
-----Original Message-----
From: Jarno(_dot_)Elovirta(_at_)nokia(_dot_)com
[mailto:Jarno(_dot_)Elovirta(_at_)nokia(_dot_)com]
Sent: Thursday, September 16, 2004 9:48 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] not getting expected matches
Hi,
okay, the expression I'm using is:
<xsl:template match="//*[starts-with(name(), 'SEG')]/NAME">
<xsl:value-of select="."/><xsl:text>*</xsl:text>
</xsl:template>
I'm thinking that this should return me the text of the
'NAME' child of all of the elements that start with 'SEG'
correct? but I have this as my XML...
No, you don't have an expression to select anything, but
rather you have a match pattern that matches a NAME element
whose parent element's name starts with "SEG".
Cheers,
Jarno
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