XSL to convert an XML in similar format's without changing any tags

[Amit]

Hi All,

I am stuck in a pretty simple problem.

I need an XSL to convert an XML with similar format's and some
additions, please consider the following example to understand my
issue.

Input XML

<Root>
<Name>amit</Name>
<Name1>gupta</Name1>
<other>bla bla bla</other>
<Root>

Output Desired XML

<Root>
<Name>amit</Name>
<Name1>gupta</Name1>
<other>bla bla bla</other>
<AdditionalTagA ddition to previous xml </AdditionalTag>
<Root>

I know, how to add new tag's. My problem is that I want some generic
XSL to convert, all tag's in input xml, no matter how many they are,
to output xml, with some additional tags.

Thanks in Advance,
Regards,
-Amit Gupta

[Martin Honnen]

Amit wrote:

<Root>
<Name>amit</Name>
<Name1>gupta</Name1>
<other>bla bla bla</other>
<Root>

Output Desired XML

<Root>
<Name>amit</Name>
<Name1>gupta</Name1>
<other>bla bla bla</other>
<AdditionalTagA ddition to previous xml </AdditionalTag>
<Root>

<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>

<xsl:template match="Root">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
<AdditionalTagA ddition to previous xml </AdditionalTag>
</xsl:copy>
</xsl:template>

should do it. If there are different kind of root elements then it
should also suffice to use e.g.

<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>

<xsl:template match="/*">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
<AdditionalTagA ddition to previous xml </AdditionalTag>
</xsl:copy>
</xsl:template>
--

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