You've supplied some code here, and some XML, presumably the input document
(though it doesn't quite match: child element color versus attribute
@colorCode). And you imply that the code doesn't do what you want it to do.
So what exactly is your problem? I can't work out from this code what your
program was intended to do, so I can't tell you why it doesn't do it.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: gi rob [mailto:primocazzuto(_at_)hotmail(_dot_)com]
Sent: 17 November 2006 18:26
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] XSLT override an existing template
I have done a working simple sample that i send ...So the
problem is in the existing xsl to be extend ....that i cannot
modify....
--------------imported.xsl------------------------------------
----------------
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/">
<xsl:apply-templates select="shirts/shirt" mode="a"/>
</xsl:template>
<xsl:template match="shirts/shirt" mode="a">
<maglia>aaaaaa</maglia>
</xsl:template>
<xsl:template match="shirts/shirt" mode="b">
<maglia>bbbbbb</maglia>
</xsl:template>
</xsl:stylesheet>
-----------------------------------------------import_ex.xsl--
-------------------------------------------------------
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:import href="imported.xsl"/>
<xsl:output method="xml" indent="yes"/>
<xsl:template match="shirts/shirt" mode="a" >
<xsl:copy-of select="."/>
</xsl:template>
<xsl:template name = "find_col">
<xsl:param name="colore" />
<shirt>
<xsl:for-each select="$colore/color[(_at_)cid='c7']">
<xsl:copy-of select="."/>
</xsl:for-each>
</shirt>
</xsl:template>
</xsl:stylesheet>
--------------------------------------------------file.xml----
----------------------------------------------------------
<shirts>
<shirt colorCode="c4"><a>oxford button-down</a></shirt>
<shirt colorCode="c1">poly blend, straight collar</shirt>
<shirt colorCode="c6">monogrammed, tab collar</shirt>
</shirts>
--------------------------------------------------------------
--------------------------------------------------------
From: "gi rob" <primocazzuto(_at_)hotmail(_dot_)com>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] XSLT override an existing template
Date: Fri, 17 Nov 2006 15:57:15 +0000
Do you have a working example? please
Thanx
Giancarlo
From: "Michael Kay" <mike(_at_)saxonica(_dot_)com>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Subject: RE: [xsl] XSLT override an existing template
Date: Fri, 17 Nov 2006 15:43:43 -0000
Yes, it's possible. If it's not working, then you've got
something wrong.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: gi rob [mailto:primocazzuto(_at_)hotmail(_dot_)com]
Sent: 17 November 2006 15:23
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] XSLT override an existing template
Hi , i m new. I wanna know if is possible to override an
existing template with the "mode" attribute specified, via
xsl:import. Because i have done a simple example working only
without "mode".
Thanx to all
Giancarlo
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