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

2008-10-07 07:12:23
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(_at_)aptaracorp(_dot_)com> 
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

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