Sorting a xsl:if test="position()"

charmaigne · April 11th, 2006, 12:38 PM

Hi All,

I am wondering if someone can help.
I am trying to sort an XSL transformation where the output is from xsl:if statements....
I need to sort the output in ascending order of the following xsl:if
<xsl:if test="position()=9">

Charmaigne
================================================== ==
<?xml version="1.0" encoding="ISO-8859-1" ?>
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:qq="urn:schemas-microsoft-com:office:spreadsheet" version="1.0">
<xsl:template match="/">
    <HTML>
    <TABLE>
   <xsl:for-each select="qq:Workbook/qq:Worksheet/qq:Table/qq:Row">
   <TR>
    <xsl:for-each select="qq:Cell">
            <xsl:if test="position()=1">
                <TD>
                    <xsl:value-of select="qq:Data"/>
                </TD>
            </xsl:if>
            <xsl:if test="position()=9">
                <TD>
                    <xsl:value-of select="qq:Data"/>
                </TD>
            </xsl:if>
        <xsl:if test="position()=7">
            <TD>
                <xsl:value-of select="qq:Data"/>
            </TD>
        </xsl:if>
        <xsl:if test="position()=11">
            <TD>
                <xsl:value-of select="qq:Data"/>
            </TD>
        </xsl:if>
        <xsl:if test="position()=4">
            <TD>
                <xsl:value-of select="qq:Data"/>
            </TD>
        </xsl:if>
    </xsl:for-each>
       </TR>
   </xsl:for-each>

    </TABLE>
    </HTML>
</xsl:template>
</xsl:stylesheet>

mhkay · April 11th, 2006, 05:39 PM

Your inner for-each seems a rather convoluted way of writing

<xsl:for-each select="qq:Cell[position() = (1,4,7,9,11)]">
                <TD>
                    <xsl:value-of select="qq:Data"/>
                </TD>
</xsl:for-each>

That's 2.0 syntax: in 1.0 you have to write the predicate as

position() = 1 or position() = 4 or ...

You don't make it clear, but I suspect you want to sort the rows (the outer for-each) not the columns. So the outer for-each becomes

<xsl:for-each select="qq:Workbook/qq:Worksheet/qq:Table/qq:Row">
  <xsl:sort select="qq:Cell[9]/qq:Data"/>

You need to be aware that there is no guarantee when you export an Excel spreadsheet as XML that the 9th Cell in a Row represents the contents of column 9 in the spreadsheet. Empty cells and cells spanning multiple columns can affect the count. But perhaps it works for your particular spreadsheet.

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

charmaigne · April 12th, 2006, 04:42 AM

Michael,

Thankyou so much, the sort worked fantastically...
The first part to the solution concating all the position() in one did not, I think this may be as you sugessted that I am using and old version on XSL...

Charmaigne