Re: [xsl] Displaying grid of elements in table with the position() call for input names?

Subject: Re: [xsl] Displaying grid of elements in table with the position() call for input names?
From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx>
Date: Thu, 8 Mar 2001 09:10:06 +0000
Hi Kevin,

> This works fine. But now, I have dilema..I need to display a random
> number of fields horizontally as well. Each row has the same number
> of fields across horizontally. So, the name should be something like
> SomeName_rowNum_colNum. However, since the position() call seems to
> be giving me the row number, what happens when I use it in this
> context..where by I am in the row loop and now building the column
> of fields. Does position() give me the number of fields in the
> current xsl:for-each loop? If so, how can I access the "outer"
> xsl:for-each loops count?

You just need to store it in a variable:

  <xsl:for-each select="$rows">
     <xsl:variable name="rowNum" select="position()" />
     <xsl:for-each select="$cols">
        <xsl:variable name="colNum" select="position()" />
        <input type="text" name="SomeName_{$rowNum}_{$colNum}" />
     </xsl:for-each>
  </xsl:for-each>

I hope that helps,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/



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