Re: [xsl] Read directory and its contents and put it to another file

["Mukul Gandhi" <gandhi.mukul@xxxxxxxxx>]

You can do something like (a modified identity transformation),

<xsl:variable name="inputFiles"
select="collection(concat($dir,'?select=*.tbl'))" />

<xsl:template match="node() | @*">
  <xsl:copy>
    <xsl:apply-templates select="node() | @*" />
  </xsl:copy>
</xsl:template>

<xsl:template match="span">
  <xsl:variable name="idVal" select="@id" />
  <xsl:copy-of select="$inputFiles/*[contains(base-uri(.), $idVal)]" />
</xsl:template>

This comes quickly out of my mind, and is not tested.

On Tue, Oct 7, 2008 at 4:19 PM, J. S. Rawat <jrawat@xxxxxxxxxxxxxx> wrote:
> Hi list
> Have anyone an idea how to read directory. I am using saxon with xsl 2.0 and
> want to do it by passing arguments
> java -jar saxon8.jar a.xml a.xsl tables=tables >b.htm
>
> 1. I have a table directory which have 'n' number of coded tables.
> 2. Second i have an xml file where tabels are called.
> 3. I have to pull the table from table directory and paste it into the
> output file.
>
> table>dir
> t1.tbl
> t2.tbl
> ...
> tn.tbl
>
> Content of 1.tbl
> <table>
> ...
> </table>
>
> Input
> <p>xxx <span id="t1"/> yyy </p>
>
> Output
> <p>xxx <table>...</table> yyy</p>
>
> Thanks in advance
> ...JSR


-- 
Regards,
Mukul Gandhi