Reformatting xml

[Riccardo]

Hello,

I have an xml sample string like this:
<?xml version="1.0" encoding="ISO-8859-1"?><catalog><cd><title>Empire
Burlesque</title><artist>Bob
Dylan</artist><country>USA</country><company>Columbia</company><price>10.90</price><year>1985</year></cd></catalog>

and I need to obtaing a new reformatted version like this for a better
display in a richtextbox control:
<?xml version="1.0" encoding="ISO-8859-1"?>
<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
</catalog>
How can I do that?
Maybe I can do that with xsl or a c# recoursive procedure but what's
better solution?

Thanks

[Martin Honnen]

Riccardo wrote:

and I need to obtaing a new reformatted version like this for a better
display in a richtextbox control:
<?xml version="1.0" encoding="ISO-8859-1"?>
<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
</catalog>
How can I do that?
Maybe I can do that with xsl or a c# recoursive procedure but what's
better solution?

Apply an XSLT stylesheet like this:

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

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

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

</xsl:stylesheet>

With .NET 1.x you should be using System.Xml.Xsl.XslTransform, with .NET
2 and 3 System.Xml.Xsl.XslCompiledTransform to run an XSLT transformation.


--

Martin Honnen --- MVP XML
http://JavaScript.FAQTs.com/

[Riccardo]

Martin Honnen ha scritto:

Apply an XSLT stylesheet like this:

great, thank you.