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39

answers:

1

Hi folks, I have a tricky XSLT transformation and I'd like your advise My xml is formatted as below:

<Person>
<name>John</name>
<date>June12</date>
<workTime taskID=1>34</workTime>
<workTime taskID=2>12</workTime>
</Person>
<Person>
<name>John</name>
<date>June12</date>
<workTime taskID=1>21</workTime>
<workTime taskID=2>11</workTime>
</Person>

The output xml should be:

<Person>
<name>John</name>
<taskID>1</taskID>
<workTime>
    <date>June12</date>
    <time>34</time>
</worTime>
<workTime>
    <date>June13</date>
    <time>21</time>
</worTime>
</Person>
<Person>
<name>John</name>
<taskID>2</taskID>
<workTime>
    <date>June12</date>
    <time>12</time>
</worTime>
<workTime>
    <date>June13</date>
    <time>11</time>
</worTime>
</Person>

Essentially, as an input, a "Person" object gathers all the task/workTime for a specific date. As an output, I want the "Person" object to gather the date/workTime for a specific task.

I need to use XLST 1.0. I've been trying to use grouping with key but get very puzzled.

Appreciate your help. Daniel

A: 

I believe one way to solve it is to use muenchian grouping, but in this case you are grouping by task and by person, and so you need an aggregated key

<xsl:key name="PersonTasks" match="workTime" use="concat(@taskID, ../name)"/>

This key is used to look up all distinct 'person tasks'. You would iterate over this by doing the following

<xsl:apply-templates select="//workTime[generate-id() = generate-id(key('PersonTasks',concat(@taskID, ../name))[1])]"/>

Then, for each distinct Person and TaskID, you just need to find all the work times. Try this stylesheet (Note I have added a 'People' element, to give the output a single root element to ensure it is valid XML)

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;

   <xsl:key name="PersonTasks" match="workTime" use="concat(@taskID, ../name)"/>

   <xsl:template match="/">
      <People>
         <xsl:apply-templates select="//workTime[generate-id() = generate-id(key('PersonTasks',concat(@taskID, ../name))[1])]"/>
      </People>
   </xsl:template>

   <xsl:template match="workTime">
      <xsl:variable name="taskID">
         <xsl:value-of select="@taskID"/>
      </xsl:variable>
      <xsl:variable name="name">
         <xsl:value-of select="../name"/>
      </xsl:variable>
      <Person>
         <name>
            <xsl:value-of select="$name"/>
         </name>
         <taskID>
            <xsl:value-of select="$taskID"/>
         </taskID>
         <xsl:for-each select="//workTime[../name = $name][@taskID = $taskID]">
            <workTime>
               <date>
                  <xsl:value-of select="../date"/>
               </date>
               <time>
                  <xsl:value-of select="."/>
               </time>
            </workTime>
         </xsl:for-each>
      </Person>
   </xsl:template>

</xsl:stylesheet>

When you apply this to the following input XML

<People>
<Person>             
<name>John</name>             
<date>June12</date>             
<workTime taskID="1">34</workTime>             
<workTime taskID="2">12</workTime>             
</Person>             
<Person>             
<name>John</name>             
<date>June13</date>             
<workTime taskID="1">21</workTime>             
<workTime taskID="2">11</workTime>             
</Person>   
</People>

You should get the following output

<People>
   <Person>
      <name>John</name>
      <taskID>1</taskID>
      <workTime>
         <date>June12</date>
         <time>34</time>
      </workTime>
      <workTime>
         <date>June13</date>
         <time>21</time>
      </workTime>
   </Person>
   <Person>
      <name>John</name>
      <taskID>2</taskID>
      <workTime>
         <date>June12</date>
         <time>12</time>
      </workTime>
      <workTime>
         <date>June13</date>
         <time>11</time>
      </workTime>
   </Person>
</People>
Tim C