Xpath Help

[samueln12]

Hi ALL,

I have a XML file in following type

<xml>
<para type="P">Sample text 1 goes here</para>
<para>Sample text 2 goes here</para>
<para>Sample text 3 goes here</para>
<para type="P">Sample text 4 goes here</para>
<para>Sample text 5 goes here</para>
<para>Sample text 6 goes here</para>
<para>Sample text 7 goes here</para>
<para type="P">Sample text 8 goes here</para>
<para type="P">Sample text 9 goes here</para>
</xml>

I want the XML in following type

<xml>
<para type="P">Sample text 1 goes here</para>

<para type="P">
<para>Sample text 2 goes here</para>
<para>Sample text 3 goes here</para>
Sample text 4 goes here</para>

<para type="P">
<para>Sample text 5 goes here</para>
<para>Sample text 6 goes here</para>
<para>Sample text 7 goes here</para>
Sample text 8 goes here</para>
<para type="P">Sample text 9 goes here</para>
</xml>

I don't know how to handle this situation. I tried using preceding-sibling and following-sibling.

Please help me out from this situation

Thanks
Sam

[jkmyoung]

Modified muenchian grouping:

<xsl:key name="paraP" match="para" use="generate-id((following-sibling::para[@type='P'])[1])"/>

<xsl:template match="xml">
<copy>
  <xsl:apply-templates match="para[@type='P']"/>
</copy>
</xsl:template>

<xsl:template match="para">
  <para type="P">
    <xsl:for-each select="key('paraP', generate-id(.))[not (@type='P')]">
      <xsl:copy-of select="."/>
    </xsl:for-each>
    <xsl:value-of select="."/>
  </para>
</xsl:template>

It looks like you're using para[@type="P"] as terminators? New lines might be off a little; modify accordingly.