I have a question regarding the parsing of command line specificed
parameters for an XSLT stylesheet - what appears to happen is my parameter
is being interpreted as a literal string instead of as an XPath expression.

My XML data is: person.xml
<?xml version="1.0" encoding="UTF-8"?>

<NewDataSet>

<Person>

<pID>1234</pID>

<LastName>Laub</LastName>

<FirstName>David</FirstName>

<Addr1>123 Jerome</Addr1>

<Addr2>Appt 10</Addr2>

<City>Skokie</City>

<State>IL</State>

<ZipCode>60076</ZipCode>

</Person>

</NewDataSet>

My XSLT stylesheet is: entity.xsl

<?xml version="1.0" encoding="UTF-8"?>

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

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

<xsl:param name="Entity" select="./NewDataSet/Person" />

<xsl:template match="NewDataSet">

<xsl:copy-of select="$Entity"/>

</xsl:template>

</xsl:stylesheet>

My MSXSL command line is ANY of the following

msxsl person.xml entity.xsl -o Entity.xml Entity=./NewDataSet/Person

msxsl person.xml entity.xsl -o Entity.xml Entity="./NewDataSet/Person"
(note double quotes)

msxsl person.xml entity.xsl -o Entity.xml Entity='./NewDataSet/Person'
(note single quotes)

The ouput of this transform is:

<?xml version="1.0" encoding="UTF-8"?>./NewDataSet/Person  (in other words
my parameter is being interpreted as a literal string instead of an XPath
expression)

I CAN get the correct output (the Person sub-tree) in XMLSPY when I specify
the parameter Entity as ./NewDataSet/Person (NO quotes)

<?xml version="1.0" encoding="UTF-8"?>

<Person>

<pID>1234</pID>

<LastName>Laub</LastName>

<FirstName>David</FirstName>

<Addr1>123 Jerome</Addr1>

<Addr2>Appt 10</Addr2>

<City>AltamoSkokie</City>

<State>IL</State>

<ZipCode>^0076</ZipCode>

</Person>

So CAN I specify my parameter as an XPath expression (versus a string
literal) with the command line MSXSL utility?

Thanks