Thanks,
I'll work through these tonight.  Appreciate your help.
Mark

Two possible solutions using identity templates. The first one uses a
dedicated mode for the nested items:

  <xsl:template match="List">
    <xsl:copy>
      <xsl:apply-templates select=".//Item"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="Item">
    <xsl:copy>
      <xsl:apply-templates mode="suppress-item"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="* | @*" mode="#all">
    <xsl:copy>
      <xsl:apply-templates select="@*, node()" mode="#current"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="List">
    <xsl:copy>
      <xsl:for-each select=".//Item">
        <xsl:copy>
          <xsl:apply-templates/>
        </xsl:copy>
      </xsl:for-each>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="* | @*">
    <xsl:copy>
      <xsl:apply-templates select="@*, node()" mode="#current"/>
    </xsl:copy>
  </xsl:template>

Both wont leave an empty ElementA though because of the contained text
nodes.

I have XML consisting of <Item> elements that have nested <Item>
elements at various depths within the tree, in its simplest form
something like:

<List>
    <Item>
        <ElementA>
            <Item>
                <ElementB/>
            </Item>
        </ElementA>
    </Item>
</List>

That I want to render as:

<List>
    <Item>
        <ElementA/>
    </Item>
    <Item>
        <ElementB/>
    </Item>
</List>

Is there a general technique I can use to remove nested <Item> elements
whatever their depth and place all <Item> elements at the same level
within the <List>?
Thanks,
Mark

--
Gerrit Imsieke
Geschdftsf|hrer / Managing Director
le-tex publishing services GmbH
Weissenfelser Str. 84, 04229 Leipzig, Germany
Phone +49 341 355356 110, Fax +49 341 355356 510
gerrit.imsieke@xxxxxxxxx, http://www.le-tex.de

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