hello

I am trying to run in for and in the same line to put what the response is. example to what I mean:

<xsl:for-each select = "$counter-db_columns[@class='statistic']">
  <case>
    <xsl:attribute name = "v2">
      <xsl:value-of select="@dbcolumn"/>
    </xsl:attribute>
  </case>
</xsl:for-each>

what I want is to know if there is a function or or some known code, that will put all the out put dbcolumn in v2, v3, v4 and so on, as many dbcolumn that there are.

thanks

the case v2,v3, v4 should be in the same line, and to not write case for each dbcolumn

Yes, I'm sure it's achievable, but I don't really understand what your source document looks like. Please provide an example of the source document and the desired output.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

source

<etd version="1.0">
 <version major="1" minor="0" />
  <counters>
   <counter tid="101">
    <display color="16" colorRGB="124" sortable="yes" sum-format="byte" avg-format="byte" />
     <data class="statistic" type="float" dbcolumn="BYTES_READ" mandatory="no" func="SUM" collection-id="bytes-read" />
   </counter>
   <counter tid="102">
    <display color="17" colorRGB="125" sortable="yes" sum-format="byte" avg-format="byte" />
    <data class="statistic" type="float" dbcolumn="BYTES_WRITE" mandatory="no" func="SUM" collection-id="bytes-write" />
   </counter>
  </counters>
</etd>

out put should look something like
<case v2="BYTES_READ" v3="BYTES_WRITE">
...
...

now each dbcolumn is different case, and since what filling the case is identical, I though to put all the cases with the same result together

thanks

Try:

<case>
  <xsl:for-each select="counter">
    <xsl:attribute name="v{position()}">
      <xsl:value-of select="@dbcolumn"/>
    </xsl:attribute>
  </xsl:for-each>
</case>

(But I don't know why your first attribute is v2?)

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

thanks

if I want the count to start from a specific number and not from v1 (
for example if I want the first v to be v3 and to start count from there?

Then replace position() by position()+2.

XPath is an expression language. Any place you can use an expression (like position()), you can use any expression (like position()+2), so long as its value is of the right type.

So this shouldn't really be very difficult to work out for yourself.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference