Re: [xsl] Xslt string Operation

["Mailing Lists Mail daktapaal@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

So,
The params had default values and that is exactly the way i am expecting
the caller to send these params. With fwd slashes, with .xml as file name
extention and always windows file name.. that is kind of why i put the
value in the param to give an example.
What i did works, but wanted to know if there was a better way.
Dak.

On Aug 24, 2016 4:13 PM, "Michael Kay mike@xxxxxxxxxxxx" <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> 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
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