xPath questions

[Raymond Law]

Would you give some advise of how to use xPath to convert attribute to element?
The original XML
<Doclists>
<Department name="IS">
<Cateogry name="MISC">
<Document>
<DocName>Mydocument</DocName>
<DocAuthor>Someone</ocAuthor>
</Document>
</Category>
</Department>
</Doclists>
I need to convert the above XML as follow:
<Doclists>
<Document>
<DocName>Mydocument</DocName>
<DocAuthor>Someone</ocAuthor>
<Departmen>IS</Department>
<Category>MISC</Category>
</Document>
</Doclists>
thanks your help in advance

[Valentin Crettaz]

rlaw,
Welcome to Javaranch
We'd like you to read the Javaranch Naming Policy and change your publicly displayed name (change it here) to comply with our unique rule. Thank you.
PS: Quote from the naming policy:

For your publicly displayed name, use a first name, a space, and a last name. Obviously fictitious names or improperly formatted names may be locked out.

[Dan Drillich]

Here is a modest start. It converts the contents of attribute name to be the department element data.

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

<xsl:template match="Department">
    <Department>
    <xsl:value-of select="@name"/>
    </Department>
    <xsl:copy-of select="Category "/>
</xsl:template>

</xsl:stylesheet>  

Cheers,
Dan

[Raymond Law]

But how to retreive parent attribute when parsing child element like following?
The following doesn't work?
<xsl:template match="Document">
<Department>
<xsl:value-of select="parent::attribute::name"/>
</Department>
</xsl:template>
</xsl:stylesheet>

I really appreicate your help!!

[Dan Drillich]

Please try -

<xsl:template match="Document"> 
<Department> 
<xsl:value-of select="../../@name"/> 
</Department> 
</xsl:template>  

Cheers,
Dan

[Raymond Law]

Dan,
My problem finally got solved. I thought that I need to use "preceding-sibling" to get parent attribute, but "../../" can easily solve my problem. I need to spend more time on xPath.
Anyway, I really appreicate your help!

Thanks you!!