Subject: Re: [xsl] Xslt string Operation From: "Michael Kay mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Wed, 24 Aug 2016 20:13:25 -0000 |
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@xxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: > > Dear all, > I have the following two params > <xsl:param name = bfileHrefb select = b bP:/developers/perf/bigPayload.xmlb b/> > <xsl:param name="outDir" select="'outDir'"/> > > I want OutFileName to be : bP:/developers/perf/outDir/bigPayload-Formatted.xmlb > > 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 > > XSL-List info and archive <http://www.mulberrytech.com/xsl/xsl-list> > EasyUnsubscribe <-list/293509> (by email <>)
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