How to tranform an XML tag value to an atrribute value

[su]

HI,

I want to tranform following
<extends>/public/a1.xml</extends>

into
<extends xlink:type="simple" xlink:href="/public/a1.xml"></extends>

using XSL.

How does the code should look like in XSL file??

Sumit

[Martin Honnen]

su wrote:

I want to tranform following
<extends>/public/a1.xml</extends>

into
<extends xlink:type="simple" xlink:href="/public/a1.xml"></extends>

using XSL.

<xsl:template match="extends">
<xsl:copy>
<xsl:attribute name="xlink:type"

namespace="http://www.w3.org/1999/xlink">simple</xsl:attribute>
<xsl:attribute name="xlink:href"

namespace="http://www.w3.org/1999/xlink"><xsl:value-of
select="."/></xsl:attribute>
</xsl:copy>
</xsl:template>

is one way that is rather generic, it might suffice to use the much shorter

<xsl:template match="extends">
<extends
xlink:type="simple"
xlink:href="{.}"
xmlns:xlink="http://www.w3.org/1999/xlink" />
</xsl:template>

--

Martin Honnen
http://JavaScript.FAQTs.com/

[su]

Hi Martin, Thanks for the solution,
In the mean time I wrote this code and it worked:-

<xsl:template match="extends">
<xsl:variable name="xdbextend">
<xsl:value-of select="."/>
</xsl:variable>
<extends xlink:href="{$xdbextend}" xlink:type="simple"
xlmns:xlink="http://www.w3.org/1999/xlink">>
</extends>

Thanks .