Re: [xsl] How to pass search path as variable and get back node list.

["J.Pietschmann" <j3322ptm at yahoo dot de>]

Andrew Kerns wrote:
> <xsl:param name="DATA_REQUEST_NAME"/>  
> 
> <xsl:variable name="ftype_search"
> select="concat('//JOB_REQUEST/DATA_REQUEST[@NAME=',$DATA_REQUEST_NAME,']/FILE_TYPE')"/>

You have assigned a string to the variable $ftype_search,
and it will stay a string, even if it looks like an XPath
expression. In most other languages, if you do
   a="1 + 1";
   b=a;
you usually don't expect that b is 2 afterwards.

Your problem is much simpler to solve, try this:
   <xsl:variable name="ftype_search"
      select="//JOB_REQUEST/DATA_REQUEST[@NAME=$DATA_REQUEST_NAME]/FILE_TYPE"/>

If you really want to pass more complicated expressions
as parameter values, like "1 + 1" or "/DATA_REQUEST[@NAME=$DATA_REQUEST_NAME]",
search the manual for your processor for an evaluate()
function, usually to be used like this:
   <xsl:variable name="ftype_search"
      select="xx:evaluate(concat('//JOB_REQUEST',$REQUEST))"/>
MSXML does not have an evaluate() function, use some
JScript in this case.

J.Pietschmann


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