xslt to transform format of xml document

Question

i have the xml in a structure such as this:

<RWFCriteria reportType="ProgramReview">
  <item id="36" name="" value="9" type="Milestone" />
  <item id="31" name="" value="9" type="Milestone" />
  <item id="33" name="" value="11" type="Milestone" />
</RWFCriteria>

and need to it be converted to:

<data>
  <release id="9">  <milestone id="36" /> <milestone id="31" /> </release>
  <release id="11"> <milestone id="33" /> </release>
</data>

what would the XSLT look like for this transformation?

Bronumski · Accepted Answer · 2010-11-19 17:33:28Z

1

You need to group the items based on their value attribute. If you are using xslt 1 you can do this using the Muenchian Method which would look something like:

  <xsl:key name="item-value" match="item" use="@value" />

  <xsl:template match="/RWFCriteria">

    <data>
      <xsl:for-each select="item[count(. | key('item-value', @value)[1]) = 1]">
        <release id="{@value}">
          <xsl:for-each select="key('item-value', @value)">
            <milestone id="{@id}" />
          </xsl:for-each>
        </release>
      </xsl:for-each>
    </data>

  </xsl:template>

edited Nov 19, 2010 at 17:33

user357812

answered Nov 19, 2010 at 17:21

Bronumski

14.3k6 gold badges54 silver badges80 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Bronumski Over a year ago

@Alejandro Thanks for fixing the attributes

user357812 Over a year ago

No problem. +1 For a correct answer. (Besides I don't like nested xsl:for-each)

Bronumski Over a year ago

@Alejandro Me neither, I would normally palm things off to their own template but I felt it was clearer.

Collectives™ on Stack Overflow

xslt to transform format of xml document

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related