Actually, you want OutFileName to be a function of fileHref and outDir, and you
are asking us to work out what that function does from one example of its input
and output. To do this properly we need to know what the possible range of
values of fileHref and outDir is: for example, is fileHref always a Windows
filename using forwards slashes to separate the parts of the path? Will it
always end in ".xml"? Or since it's a param, might someone pass in a filename
using backslashes, with no file extension?
I would encourage you to use URIs rather than filenames. You could then use the
resolve-uri() function for at least part of the task.
Michael Kay
Saxonica
On 24 Aug 2016, at 17:16, Mailing Lists Mail daktapaal(_at_)gmail(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Dear all,
I have the following two params
<xsl:param name = “fileHref” select = “ ‘P:/developers/perf/bigPayload.xml’
”/>
<xsl:param name="outDir" select="'outDir'"/>
I want OutFileName to be :
‘P:/developers/perf/outDir/bigPayload-Formatted.xml’
Is there a better way to do this than what I did?
<xsl:variable name="OutFileName" select=" substring-before($fileHref,
tokenize($fileHref,'/')[last() -1 ] ) || $outDir || '/' ||
substring-before( tokenize($fileHref,'/')[last()] ,'.xml') ||
'-Formatted.xml'"/>
thanks.
Dt
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