Re: [xsl] Transformation of generic spreadsheet XML

Subject: Re: [xsl] Transformation of generic spreadsheet XML
From: Xiaocun Xu <xiaocunxu@xxxxxxxxx>
Date: Thu, 19 Apr 2001 13:19:52 -0700 (PDT)
Hi, Jeni:

  Very much appreciated for the well detailed answer

> To answer your question, you can get the relevant 
> cell by comparing its column
> attribute to the column attribute of the current
> cell, which you can get with:
>   current()/@column

This is exactly what I needed.

> I think you'd be better off storing the bidHeader
> row in a global
> variable (assuming that the row elements are
> children of a 'rows'
> document element here):
> <xsl:variable name="bidHeader" select="/rows/row[1]"
> />

Sorry I did not provide the whole picture in my
original post (it was already pretty long :().  The
actual source XML had many rows before the $bidHeader
row, therefore the bid header row does not have exact
position among its siblings.  So my XSL had to catch
that bid header row and process all following siblings
as follows:

<xsl:template match="row" mode="createBids">
    <xsl:element name="createBids">
select="following-sibling::row" mode="BidsDetail"/>

<xsl:template match="row" mode="BidsDetail">
  <xsl:if test="string(cell[1]) = string('BidType') or
substring(cell[1],2) = string('BidType')">
    <xsl:variable name="bidHeader" select="."/>


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