tricky xsl transformation needed...

[stefan.oedenkoven@googlemail.com]

Hi ng,

how can i transform this source-xml in the target xml? is it possible
to get <a>Ax</aenclosed?

source-xml:
<z>
A1
<b>B1</b>
A2
</z>


desired target-xml:
<z>
<a>A1</a>
<b>B1</b>
<a>A2</a>
</z>


....anyone?

[Martin Honnen]

Re: tricky xsl transformation needed...

stefan.oedenkov en@googlemail.c om wrote:

how can i transform this source-xml in the target xml? is it possible
to get <a>Ax</aenclosed?
>
source-xml:
<z>
A1
<b>B1</b>
A2
</z>
>
>
desired target-xml:
<z>
<a>A1</a>
<b>B1</b>
<a>A2</a>
</z>

<xsl:styleshe et
xmlns:xsl="http ://www.w3.org/1999/XSL/Transform"
version="1.0">

<xsl:output method="xml" indent="yes"/>

<xsl:template match="z/text()[starts-with(normalize-space(.), 'A')]">
<a>
<xsl:value-of select="normali ze-space(.)"/>
</a>
</xsl:template>

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

</xsl:stylesheet>


--

Martin Honnen

http://JavaScript.FAQTs.com/

[stefan.oedenkoven@googlemail.com]

Re: tricky xsl transformation needed...

Hi Martin,

thanks a lot...!!!

//stefan