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Hello,
It would be nice if someone could tell me the easiest way of changing just one (and keeping the other) attributes of an copied element. I am using the following templates to copy elements from a source class to a target class: <xsl:template match="node() | @*" mode="transform">
<xsl:copy>
<xsl:apply-templates select="node() | @*" mode="transform"/>
</xsl:copy>
</xsl:template>
<xsl:template match="//class[@name=$TargetCName]" mode="transform">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:copy-of select="//class[@name=$SrcCName]/field">
<xsl:apply-templates mode="transform"/>
</xsl:copy>
</xsl:template>Inside "class" are elements like <field id="31" class="3" type="type(basic,int,0)" name="publ_int_j" init="null"/> which are going to be copied. What, if I want to change the "id" of the copied element? (Suppose, I don't know that the element name is "field") Do I have to create a completely new element using <xsl:element> and to write down every single attribute explicitely using <xsl:attribute>? Or is there a possibility to copy the old attributes und to create just one new attribute? My problem is, that <xsl:copy-of> doesn't allow <xsl:attribute> inside and <xsl:copy> which allows it doesn't allow a "select" inside... Regards, Garvin
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