Converting CSV to XML without hardcoding schema detail

Hello,

I am trying to convert a CSV datafile into XMl format.
The headers for the CSV data are in a file header.csv e.g. Field1,Field2
The data is in a file Data.csv e.g.
Value11,Value12
Value21,Value22

I need to convert the CSV data into xml output by creating xml elements
using the names in the csv header and taking the corresponding values from
the data file, so that I get an xml as follows:

<doc>
<line>
<Field1>Value11</Field1>
<Field2>Value12</Field2>
</line>
<line>
<Field1>Value21</Field1>
<Field2>Value22</Field2>
</line>
</doc>

I was trying to see if I can do this without hardcoding the header names in
the xsl. I reached upto the point where my xsl looks as below:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:op="http://www.w3.org/2001/12/xquery-operators"
    xmlns:xf="http://www.w3.org/2001/12/xquery-functions" version="2.0">

    <xsl:output  name="xmlFormat" method="xml" indent="yes"
omit-xml-declaration="yes"/>

    <xsl:variable name="source1" select="'data.csv'"/>
    <xsl:variable name="elementNamesList" select="'Header.csv'"/>
    <xsl:variable name="encoding" select="'iso-8859-1'"/>

    <xsl:variable name="elementNames"
select="tokenize(unparsed-text($elementNamesList,$encoding),',')"/>
    <xsl:variable name="src">
        <doc>
            <xsl:for-each
select="tokenize(unparsed-text($source1,$encoding), '\r?\n')">
                <line>
                    <xsl:for-each select="tokenize(., ',')">
                        &lt;<xsl:value-of
select="op:item-at($elementNames,index-of(?parent of current
node?,.))"/>&gt;
                            <xsl:value-of select="."/>
                            &lt;/<xsl:value-of
select="item-at($elementNames,3)"/>&gt;
                    </xsl:for-each>
                </line>
            </xsl:for-each>
        </doc>
    </xsl:variable>

    <xsl:template match="/">
        <xsl:result-document format = "xmlFormat" href = "src1.xml">
            <xsl:copy-of select="$src"/>
        </xsl:result-document>
    </xsl:template>

</xsl:stylesheet>

In the yet-incomplete statement <xsl:value-of
select="op:item-at($elementNames,index-of(?parent of current node?,.))"/>, I
am trying to generate an xml element with the Nth field name from the
headers name list for the Nth field value. Couple of issues/questions here:

- I am getting the error "Cannot find a matching 2-argument
function named {http://www.w3.org/2001/12/xquery-operators}item-at()" when I
try to validate the xsl. What could be the reason?

- How can I get the ?parent of current node? Needed to compute the
index of the current data in the data record?

- Is there any other better way to do it? Any way that I can do the
same using xsl:element?

In general, is this the only/best way or is there any other better way to
achieve the same goal?


Thanks and Regards,

Vish.