[abbouh]
my xsl file is:
<xsl:choose>
<xsl:when test=".=list">
<xsl:variable name="n" select="12"/>
</xsl:when>
<xsl:otherwise>
<xsl:variable name="n" select="2"/>
</xsl:otherwise>
</xsl:choose>
-------------------
<xsl:variable name="m" select="$n"/>
i receive the error :invalid xpath expression!
so what the problem in <xsl:variable name="m" select="$n"/>?
The variable $n only exists in the scope of the block in which it is
defined. It is not known outside that block. So it is not known
outside the xsl:othewise block. You just have to put the whole
xsl:choose block inside the variable declaration. Then the variable
will be visible.
I think you meant select="'12'", did you not? Otherwise select will be
looking for a node named 12, and there can not be such a node, since it
is an illegal name for an xml element..
Cheers,
Tom P
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list