Some advice needed on grouping again

wokoman · July 7th, 2011, 12:00 PM

Hi there, I've looked through some of the grouping topics here, but wasn't able to find a real match so i'm hoping for some advice on what I'm doing wrong.

Here's my XML file :

Code:

<Book>
	<File>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="L3">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="L1">
			<Content>some data</Content>
		</Para>
		<Para tag="B_bold">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="B_bold">
			<Content>some data</Content>
		</Para>
	</File>
</Book>

And what I'f like to get is the following ; All Para elements having BSub tag attribute should be renamed to 'li' and grouped in a ul, the rest remains unchanged for now. So below is the outcome I'd like to get :

Code:

<Book>
	<File>
		<ul>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
		</ul>
		<Para tag="L3">
			<Content>some data</Content>
		</Para>
		<ul>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
		</ul>
		<Para tag="L1">
			<Content>some data</Content>
		</Para>
		<Para tag="B_bold">
			<Content>some data</Content>
		</Para>
		<ul>
			<li tag="BSub">
				<Content>some data</Content>
			</li>
		</ul>
		<Para tag="B_bold">
			<Content>some data</Content>
		</Para>
	</File>
</Book>

The xslt I use is the following :

Code:

	
<xsl:template match="File">
    <xsl:for-each-group select="Para" group-by="@tag = 'BSub'">
      <xsl:choose>
        <xsl:when test="current-grouping-key()">
          <ul>
			  <xsl:apply-templates select="current-group()"/>
          </ul>
        </xsl:when>
        <xsl:otherwise>
			  <xsl:apply-templates select="current-group()"/>
        </xsl:otherwise>
      </xsl:choose>
    </xsl:for-each-group>
  </xsl:template>

But this results in following : All my BSub's are grouped in a single node, and all the rest is grouped in a second node. While what I wanted is more like

> Start at first node, as soon as you see BSub's group these till the next none BSub, and continue like this till you reach the end. For some reason I'm not able to grasp the logic behind grouping, and all example on the net are rather simplified. So, if you could tell me how to do it right (and even more important why I need to do it like that) I'd be very grateful.

Code:

<Book>
	<ul>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
		<Para tag="BSub">
			<Content>some data</Content>
		</Para>
	</ul>
	<Para tag="L3">
		<Content>some data</Content>
	</Para>
	<Para tag="L1">
		<Content>some data</Content>
	</Para>
	<Para tag="B_bold">
		<Content>some data</Content>
	</Para>
	<Para tag="B_bold">
		<Content>some data</Content>
	</Para>
</Book>

Martin Honnen · July 7th, 2011, 12:04 PM

Use

Code:

<xsl:for-each-group select="*" group-adjacent="@tag = 'BSub'">
  <!-- here you can put the code you have in your posted for-each-group -->
</xsl:for-each-group>

wokoman · July 8th, 2011, 05:23 AM

Thanks Martin, this does the trick indeed. I do have a small additional question linked to grouping. Does the group-adjacent (or grouping key in general) allow for an 'OR' type of selection ? In my original XML I have several parameters that need to be transformed to the same type, so what I currently do is the following : (simplified xslt)

Code:

<xsl:for-each-group select="current-group()" group-adjacent="@tag = 'BSubMak.bull'">
<xsl:choose>
<xsl:when test="current-grouping-key()">
<ul class="std">
<xsl:apply-templates select="current-group()" mode="List"/>
</ul>
</xsl:when>

<xsl:otherwise>
<xsl:for-each-group select="current-group()" group-adjacent="@tag = 'BMak.bull'">
<xsl:choose>
<xsl:when test="current-grouping-key()">
<ul class="std">
<xsl:apply-templates select="current-group()" mode="List"/>
</ul>
</xsl:when>

<xsl:otherwise>
<!-- next parameter based group and so on ... -->

Where I'd like to get it a bit simpler as I end up with a lot of nestings. Following obviously doesn't work but is shows what I'd like to get

Code:

<xsl:for-each-group select="current-group()" group-adjacent="@tag = 'BSubMak.bull' | @tag = 'BMak.bull'">

so is there any way to work with an or statement as a key or is nesting the only option ?

Thanks again for any advice on this !

Martin Honnen · July 8th, 2011, 08:15 AM

The boolean "or" operator in XPath is named "or" and not "|", as "|" is the union operator on nodes. So doing

Code:

group-adjacent="@tag = 'BSubMak.bull' or @tag = 'BMak.bull'"

or maybe

Code:

group-adjacent="@tag = ('BSubMak.bull', 'BMak.bull')"

might do what you want.