RE: [xsl] Converting CSV to XML without hardcoding schema details in xsl

Can anybody suggest how to convert CSV data in the format

Field1,Field2
Value11,Value12

to xml like 

<Field1>Value11</Field1>
<Field2>Value12</Field2>

without hardcoding the fieldnames in the xsl?


<xsl:variable name="lines" as="xs:string*"
              select="tokenize(unparsed-text($input-file, '\r?\n'"))"/>
<xsl:variable name="field-names as="xs:string*"
              select="tokenize($lines[1], ',')"/>
<xsl:for-each select="subsequence($lines,2)">
<row>
  <xsl:variable name="cells" select="tokenize(., ',')"/>
  <xsl:for-each select="$cells">
    <xsl:variable name="p" as="xs:integer" select="position()"/>
    <xsl:element name="$fields[$p]"/>
      <xsl:value-of select="."/>
    </
  </
</
</

Michael Kay
http://www.saxonica.com/


I was thinking of something like

<xsl:for-each select="tokenize(., ',')"> &lt;<xsl:value-of 
select="item-at($elementNames,index-of(?parent of current 
node?,.))"/>&gt; <xsl:value-of select="."/> 
&lt;/<xsl:value-of 
select="item-at($elementNames,index-of(?parent of current 
node?,.))"/>&gt; </xsl:for-each>

where elementNames is a tokenized list of the fieldnames - 
but I am unable to get it to work.

-----Original Message-----
From: Pantvaidya, Vishwajit
Sent: Wednesday, June 21, 2006 12:17 AM
To: 'xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com'
Subject: [xsl] Converting CSV to XML without hardcoding

schema details

in xsl

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.

--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: 
<mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--


--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: 
<mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--



--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--