CODEWITHSUNDEEP
htmlxmlxslt
Matthew Dresser
Matthew Dresser

Reputation: 11442

Use xslt to convert xml list into html table with multiple columns

Say I have an xml document consisting of a list as shown below:

<Items>
  <Item>First Item</Item>
  <Item>Second Item</Item>
  <Item>Third Item</Item>
  <Item>4</Item>
  <Item>Five</Item>
</Items>

I want to convert it to a html table with two columns for the Item elements (I'm not fussed at the moment whether it's ordererd top-bottom-left-right or left-right-top-bottom).

<table>
  <tr>
    <td>First Item</td>
    <td>Second Item</td>
  </tr>
  <tr>
    <td>Third Item</td>
    <td>4</td>
  </tr>
  <tr>
    <td>Five</td>
    <td></td>
  </tr>
</table>

I understand I can get a table with a single column with the following xslt transform, but can't figure out how to get multiple columns.

<table>
  <xsl:for-each select="Items">
    <tr>
      <td><xsl:value-of select="Item"/></td>
    </tr>
  </xsl:for-each>
</table>

Upvotes: 0

Views: 4692

Answers (1)

Rubens Farias
Rubens Farias

Reputation: 57936

Try this:

<xsl:stylesheet version="1.0" 
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/Items">
    <table>
      <xsl:for-each select="Item[position() mod 2 = 1]">
        <xsl:variable name="pos" select="position() - 1" />
        <tr>
          <td><xsl:value-of select="."/></td>
          <td><xsl:value-of select="//Item[position() = ($pos * 2) + 2]"/></td>
        </tr>
      </xsl:for-each>
    </table>
  </xsl:template>
</xsl:stylesheet>

Upvotes: 1

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