Subject: [xsl] Re: Re: How to pass search path as variable and get back node list. From: Dimitre Novatchev <dnovatchev@xxxxxxxxx> Date: Wed, 27 Mar 2002 23:26:58 -0800 (PST) |
"J.Pietschmann" <j3322ptm at yahoo dot de> wrote: > 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 Instead of looking for a vendor:evaluate() function, a better approach is to use DOM and change a skeleton stylesheet before the transformation is applied, by setting the value of the "select" attribute of a certain global variable to the string XPath expression that is necessary to evaluate. This is basically what the XPath Visualizer does all the time. Cheers, Dimitre Novatchev. __________________________________________________ Do You Yahoo!? Yahoo! Movies - coverage of the 74th Academy Awards® http://movies.yahoo.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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