Re: [xsl] Copy XPATH

["Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

On 03/02/2025 23:32, Vasu Chakkera vasucv@xxxxxxxxx wrote:
> will this work?


That depends on your requirements you would know better than we do. It
all depends on how you want to process the "xpath" attribute value.


> <xsl:stylesheet version="3.0"
> B  B  xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> B  B  xmlns:x="http://www.w3.org/2005/xpath-functions";
> B  B  exclude-result-prefixes="x">
>
> B  B  <!-- Template to match all elements -->
> B  B  <xsl:template match="*">
> B  B  B  B  <xsl:element name="{local-name()}">
> B  B  B  B  B  B  <!-- Add title attribute if the element has a title child
-->
> B  B  B  B  B  B  <xsl:if test="tl">
> B  B  B  B  B  B  B  B  <xsl:attribute name="title">
> B  B  B  B  B  B  B  B  B  B  <xsl:value-of select="tl"/>
> B  B  B  B  B  B  B  B  </xsl:attribute>
> B  B  B  B  B  B  </xsl:if>
> B  B  B  B  B  B  <!-- Generate the XPath using the path() function and
> remove Q{} -->
> B  B  B  B  B  B  <xsl:attribute name="xpath">
> B  B  B  B  B  B  B  B  <xsl:value-of select="replace(path(), 'Q\{\}',
'')"/>
> B  B  B  B  B  B  </xsl:attribute>
> B  B  B  B  B  B  <!-- Copy the id attribute if present -->
> B  B  B  B  B  B  <xsl:if test="@id">
> B  B  B  B  B  B  B  B  <xsl:attribute name="id">
> B  B  B  B  B  B  B  B  B  B  <xsl:value-of select="@id"/>
> B  B  B  B  B  B  B  B  </xsl:attribute>
> B  B  B  B  B  B  </xsl:if>
> B  B  B  B  B  B  <!-- Copy other attributes except id -->
> B  B  B  B  B  B  <xsl:apply-templates select="@*[name() != 'id']"/>
> B  B  B  B  B  B  <!-- Apply templates to child elements -->
> B  B  B  B  B  B  <xsl:apply-templates/>
> B  B  B  B  </xsl:element>
> B  B  </xsl:template>
>
> B  B  <!-- Template to copy attributes -->
> B  B  <xsl:template match="@*">
> B  B  B  B  <xsl:attribute name="{name()}">
> B  B  B  B  B  B  <xsl:value-of select="."/>
> B  B  B  B  </xsl:attribute>
> B  B  </xsl:template>
>
> </xsl:stylesheet>