There is also local-name() if for some reason you want to do get the
name without a namespace prefix. I could be misunderstanding how
local-name() works compared to name(), but hopefully someone else can
clarify if I am wrong about that.
Eric
On 6/6/07, Michael Kay <mike(_at_)saxonica(_dot_)com> wrote:
You can get the name of a node using the XPath name() function, for example
<attribute type="{name()}"/>
Michael Kay
http://www.saxonica.com/
> -----Original Message-----
> From: Chaudhary, Harsh [mailto:HCHAUDHA(_at_)amfam(_dot_)com]
> Sent: 06 June 2007 22:33
> To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
> Subject: [xsl] How to get element name for output
>
> Hi,
>
> I am writing XSLt (ver 1.0) for output to an XML file. At one
> point I would like to get the name of the element and include
> that in the output. An example would be:
>
> Input XML:
>
> <rootElement>
> <productAttributeList>
> <attributeMap>
> <ReinstatementInProc>N</ReinstatementInProc>
> </attributeMap>
> <elementName>attribute</elementName>
> </productAttributeList>
> </rootElement>
>
> Output XML:
>
> <rootElement>
> <attribute type="ReinstatementInProc">N</attribute>
> </rootElement>
>
> I would really appreciate your help on this.
>
> Thanks,
> Harsh.
>
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