views:

111

answers:

3

Okay guys I think this is a tricky one...

I am looking for a clean way to do the following using XSLT:

Convert this source:

<para>blah blah</para>
<list>num1</list>
<list>num2</list>
<list>num3</list>
<para>blah blah</para>
<list>num1</list>
<list>num2</list>
<para>blah blah blah blah blah</para>

To this output:

<p>blah blah</p>
<ol>
<li>num1</li>
<li>num2</li>
<li>num3</li>
</ol>
<p>blah blah</p>
<ol>
<li>num1</li>
<li>num2</li>
</ol>
<p>blah blah blah blah blah</p>

Keep in mind I do not know exactly how many <list>'s there will be.

So far I have this:

<xsl:template match="para">
<p><xsl:value-of select="." /></p>
</xsl:template>

<xsl:template match="list">
<ol><li><xsl:value-of select="." /></li></ol>
</xsl:template>

But my output looks like this:

<p>blah blah</p>    
<ol><li>num1</li></ol>
<ol><li>num2</li></ol>
<ol><li>num3</li></ol>
<p>blah blah</p>
<ol><li>num1</li></ol>
<ol><li>num2</li></ol>
<p>blah blah blah blah blah</p>

I know why I am getting duplicate <ol> elements, but I do not know how to stop it. Quite a brain teaser.

Any help would be greatly appreciated.

+1  A: 

This transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:key name="kFollowing" match="list"
  use="generate-id(preceding-sibling::para[1])"/>

 <xsl:template match="node()|@*" name="identity">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="para">
  <p>
    <xsl:apply-templates/>
  </p>
   <xsl:variable name="vFol"
    select="key('kFollowing',generate-id())"/>
   <xsl:if test="$vFol">
      <ol>
       <xsl:apply-templates mode="copy"
           select="key('kFollowing',generate-id())"/>
      </ol>
  </xsl:if>
 </xsl:template>

 <xsl:template match="list" mode="copy">
  <li><xsl:value-of select="."/></li>
 </xsl:template>
 <xsl:template match="list"/>
</xsl:stylesheet>

when applied on the following XML document (wrapping the provided input in a single top element):

<t>
    <para>blah blah</para>
    <list>num1</list>
    <list>num2</list>
    <list>num3</list>
    <para>blah blah</para>
    <list>num1</list>
    <list>num2</list>
    <para>blah blah blah blah blah</para>
</t>

produces the wanted, correct result:

<t>
    <p>blah blah</p>
    <ol>
        <li>num1</li>
        <li>num2</li>
        <li>num3</li>
    </ol>
    <p>blah blah</p>
    <ol>
        <li>num1</li>
        <li>num2</li>
    </ol>
    <p>blah blah blah blah blah</p>
</t>

UPDATE: The OP has indicated in a comment that now he wants a solution where any non-list element can delimit a group of adjacent list siblings.

Here is the solution to the changed question :

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:key name="kFollowing" match="list"
  use="generate-id(preceding-sibling::*[not(self::list)][1])"/>

 <xsl:template match="node()|@*" name="identity">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="*[not(self::list) and following-sibling::*[1][self::list]]">
  <xsl:call-template name="identity"/>
   <xsl:variable name="vFol"
    select="key('kFollowing',generate-id())"/>
   <xsl:if test="$vFol">
      <ol>
       <xsl:apply-templates mode="copy"
           select="key('kFollowing',generate-id())"/>
      </ol>
  </xsl:if>
 </xsl:template>

 <xsl:template match="list" mode="copy">
  <li><xsl:value-of select="."/></li>
 </xsl:template>
 <xsl:template match="list"/>
</xsl:stylesheet>

When this transformation is applied on the following XML document (Note that the separating elements have now random names):

<t>
    <bara>blah blah</bara>
    <list>num1</list>
    <list>num2</list>
    <list>num3</list>
    <vara>blah blah</vara>
    <list>num1</list>
    <list>num2</list>
    <dara>blah blah blah blah blah</dara>
</t>

the wanted, correct result is produced:

<t>
    <bara>blah blah</bara>
    <ol>
        <li>num1</li>
        <li>num2</li>
        <li>num3</li>
    </ol>
    <vara>blah blah</vara>
    <ol>
        <li>num1</li>
        <li>num2</li>
    </ol>
    <dara>blah blah blah blah blah</dara>
</t>
Dimitre Novatchev
+2  A: 

This XSLT 1.0 stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:template match="node()">
        <xsl:apply-templates select="node()[1]|following-sibling::node()[1]"/>
    </xsl:template>
    <xsl:template match="para">
        <p>
            <xsl:value-of select="."/>
        </p>
        <xsl:apply-templates select="following-sibling::node()[1]"/>
    </xsl:template>
    <xsl:template match="list[preceding-sibling::node()[1][not(self::list)]]">
        <ol>
            <xsl:call-template name="makeList"/>
        </ol>
        <xsl:apply-templates select="following-sibling::node()
                                              [not(self::list)][1]"/>
    </xsl:template>
    <xsl:template match="list" name="makeList">
        <li>
            <xsl:value-of select="."/>
        </li>
        <xsl:apply-templates select="following-sibling::node()[1]
                                              [self::list]"/>
    </xsl:template>
</xsl:stylesheet>

Output:

<p>blah blah</p>
<ol>
        <li>num1</li>
        <li>num2</li>
        <li>num3</li>
</ol>
<p>blah blah</p>
<ol>
        <li>num1</li>
        <li>num2</li>
</ol>
<p>blah blah blah blah blah</p>

Note: Fine grained traversal.

Edit: Compact code.

Alejandro
Thanks for the help this is a possible approach. But isn't this approach hard to maintain? Because I will have to add the following-sibling select on every top level element I match on.
joe
@joe: This stylesheet has four rules: a default fine grained traversal, a rule for `p`, a rule for `ol` and a rule for `li`. So, it's a one to one transformation. I don't see the *hard to maintain* problem. You've asked to Dimitre: *what if I have an element H1 before the list*. Well, you have to declare waht to do with `H1` then. Now, it's bypassed. So, two solutions: copy everything you don't want to transform or add a rule for `H1`.
Alejandro
What I am saying is if I add <xsl:template match="Heading1"><h1> <xsl:value-of select="."/></h1></xsl:template>. I will have to put <xsl:apply-templates select="following-sibling::node()[1]"/> in the match as well.
joe
@joe: Yes, applying templates to first child and first following sibling is the way this works.
Alejandro
+2  A: 

XSLT 2.0 has tools especially for this kind of Operations:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:template match="xml">
        <xsl:for-each-group select="*" group-adjacent="boolean(self::list)">
            <xsl:choose>
                <xsl:when test="current-grouping-key()">
                    <ol>
                        <xsl:apply-templates select="current-group()"/>
                    </ol>
                </xsl:when>
                <xsl:otherwise>
                    <xsl:apply-templates select="current-group()"/>
                </xsl:otherwise>
            </xsl:choose>
        </xsl:for-each-group>
    </xsl:template>
    <xsl:template match="para">
        <p>
            <xsl:apply-templates/>
        </p>
    </xsl:template>
    <xsl:template match="list">
        <li>
            <xsl:apply-templates/>
        </li>
    </xsl:template>
</xsl:stylesheet>

With this XML:

<xml>
    <para>blah blah</para>
    <list>num1</list>
    <list>num2</list>
    <list>num3</list>
    <para>blah blah</para>
    <list>num1</list>
    <list>num2</list>
    <para>blah blah blah blah blah</para>
</xml>

You'll get the desired Output:

<p>blah blah</p>
<ol>
    <li>num1</li>
    <li>num2</li>
    <li>num3</li>
</ol>
<p>blah blah</p>
<ol>
    <li>num1</li>
    <li>num2</li>
</ol>
<p>blah blah blah blah blah</p>

You should read up on for-each-group at http://www.w3.org/TR/xslt20/#xsl-for-each-group

Tobias Klevenz
Thanks! This looks like a great way to do it.
joe
@Tobias Klevenz: +1 good answer. But I think it's no good to use such "brick" template rules.
Alejandro
@joe: XSLT 1.0 and XSLT 2.0 are very diferent. Please, next time tag your question as `xslt-2.0`. Also, now it's definitely a duplicate.
Alejandro