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Answer 1

A:

George, you are getting 2 rows for sub assembly as there are two elements for sub-assembly with different ids (id21 and id18) and out of the four Part15s, two belong to id21 and two belong to id18.

Since you are grouping according to the parent id, your output seems correct. If you want to have both the sub assembly elements grouped, then you need to specify the parent title rather than parent id in your concatenated key.

I just tweaked your xslt to see the grouping. This is how they are grouped. As you can see the Parent Ref for the 2nd and 3rd rows are different

Instance Ref:#id41 Parent Ref: Parent:  TopLevelAsy  1 
Instance Ref:#id15 Parent Ref:#id18 Parent: SubAsy Part15  2 
Instance Ref:#id15 Parent Ref:#id21 Parent: SubAsy Part15  2 
Instance Ref:#id19 Parent Ref:#id7 Parent: Top Assembly Part19  2 
Instance Ref:#id15 Parent Ref:#id7 Parent: Top Assembly Part15  2 
Instance Ref:#id48 Parent Ref:#id7 Parent: Top Assembly Sub Assembly  2

Edit

This is not a complete solution but you might have to do something on these lines:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
  <xsl:output method="html" indent="no"/>
  <!--   <xsl:key name="byref" match="Occurrence" use="@instancedRef"/> -->
  <xsl:key name="bytitle" match="Occurrence" use="data/title"/>
  <xsl:template match="/">
    <table border="1">
      <xsl:for-each select="DOCUMENT/ProductView/Occurrence[generate-id(.)=generate-id(key('bytitle', data/title)[1])]">
        <xsl:sort select="@id"/>
        <xsl:variable name="title" select="data/title" />
        <br />
        Title:<xsl:value-of select="$title"/>
        <xsl:variable name="driver" select="//DOCUMENT/ProductView/Occurrence[data/title = $title]/@id"/>
        Driver:<xsl:value-of select="$driver/."/>
        <!--<xsl:for-each select ="$driver">
          <xsl:value-of select="."/>
        </xsl:for-each>-->
        <xsl:for-each select="//DOCUMENT/ProductView/Occurrence[substring-after(@parentRef,'#') = $driver/.]">
          Child:<xsl:value-of select="data/title"/>
        </xsl:for-each>
      </xsl:for-each>
    </table>
  </xsl:template>
</xsl:stylesheet>

Rashmi Pandit 2009-12-11 04:42:11

When I try your suggestion using parent title, I get the following result...Parent: TopLevelAsy 1Parent: Top Assembly Part19 2Parent: Top Assembly Part15 6Parent: Top Assembly Sub Assembly 2However I need the following outputParent: TopLevelAsy 1Parent: Top Assembly Part19 2Parent: Top Assembly Part15 2Parent: Sub Asy Part15 4Parent: Top Assembly Sub Asy 2

2009-12-11 05:03:39

That's because the title that you are using is not the parent title, but its the title of the current node.

Rashmi Pandit 2009-12-11 05:10:06

In your case, SubAssembly #id18 and SubAssembly #id21 are different instances and ideally you should not group them. If they are not different but the same, then you should have only one instance in the xml and make sure that all part15 use a single parent id. On the other hand, if it is correct that they are different instances, then why do you want to group them? As per me, your xslt is quite right.

Rashmi Pandit 2009-12-11 05:13:35

Rashmi, I already got the above result and I am aware that the grouping you showed is the result of the different parent ref. Thats the root of my problem. My problem still remains...I need to group them as belowInstance Ref:#id41 Parent: TopLevelAsy 1 Instance Ref:#id15 Parent: SubAsy Part15 4 Instance Ref:#id19 Parent: Top Assembly Part19 2 Instance Ref:#id15 Parent: Top Assembly Part15 2 Instance Ref:#id48 Parent: Top Assembly Sub Assembly 2

2009-12-11 05:14:46

Answer 2

A:

George,

Here's the solution:

You need to use 2 xslts for the same. The first xslt will generate an easily parseable xml. The second xslt will use this xml as its input to achieve the desired result.

Xslt1:

Your xml will be the input to this xslt

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
  <xsl:output method="xml" indent="yes"/>
  <xsl:key name="bytitle" match="Occurrence" use="data/title"/>
  <xsl:template match="/">
    <root>

      <xsl:for-each select="DOCUMENT/ProductView/Occurrence[not(@parentRef)]">
        <xsl:element name="Data">
          <xsl:attribute name="parentTitle" />
          <xsl:attribute name="childTitle">
            <xsl:value-of select="data/title"/>
          </xsl:attribute>
        </xsl:element>
      </xsl:for-each>

      <xsl:for-each select="DOCUMENT/ProductView/Occurrence[generate-id(.)=generate-id(key('bytitle', data/title)[1])]">
        <xsl:sort select="@id"/>

        <xsl:variable name="title" select="data/title" />
        <xsl:variable name="driver" select="//DOCUMENT/ProductView/Occurrence[data/title = $title]/@id"/>

        <xsl:for-each select="//DOCUMENT/ProductView/Occurrence[substring-after(@parentRef,'#') = $driver/.]">

          <xsl:element name="Data">
            <xsl:attribute name="parentTitle">
              <xsl:value-of select="$title"/>
            </xsl:attribute>
            <xsl:attribute name="childTitle">
              <xsl:value-of select="data/title"/>
            </xsl:attribute>
          </xsl:element>

        </xsl:for-each>        
      </xsl:for-each>
    </root>
  </xsl:template>
</xsl:stylesheet>

Output of this xslt:

<?xml version="1.0" encoding="utf-8"?>
<root>
  <Data parentTitle="" childTitle="TopLevelAsy" />
  <Data parentTitle="Sub Assembly" childTitle="Part15" />
  <Data parentTitle="Sub Assembly" childTitle="Part15" />
  <Data parentTitle="Sub Assembly" childTitle="Part15" />
  <Data parentTitle="Sub Assembly" childTitle="Part15" />
  <Data parentTitle="TopLevelAsy" childTitle="Part19" />
  <Data parentTitle="TopLevelAsy" childTitle="Part15" />
  <Data parentTitle="TopLevelAsy" childTitle="Part19" />
  <Data parentTitle="TopLevelAsy" childTitle="Part15" />
  <Data parentTitle="TopLevelAsy" childTitle="Sub Assembly" />
  <Data parentTitle="TopLevelAsy" childTitle="Sub Assembly" />
</root>

Xslt2:

The output of xslt 1 should be the input to this xslt

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
  <xsl:output method="html" indent="yes"/>
  <xsl:key name="bytitle" match="Data" use="concat(@parentTitle, @childTitle)"/>
  <xsl:template match="/">
    <table border="1">
      <xsl:for-each select="/root/Data[generate-id(.)=generate-id(key('bytitle', concat(@parentTitle, @childTitle))[1])]">
        <xsl:sort select="@parentTitle"/>
        <xsl:variable name="parentTitle" select="@parentTitle"/>
        <xsl:variable name="childTitle" select="@childTitle"/>
        <tr>
          <td>
            <xsl:value-of select="@parentTitle"/>
          </td>
          <td>
            <xsl:value-of select="@childTitle"/>
          </td>
          <td>
            <xsl:value-of select="count(//root/Data[@parentTitle=$parentTitle and @childTitle = $childTitle])"/>
          </td>
        </tr>
      </xsl:for-each>
    </table>
  </xsl:template>
</xsl:stylesheet>

Output:

 TopLevelAsy 1 
Sub Assembly Part15 4 
TopLevelAsy Part19 2 
TopLevelAsy Part15 2 
TopLevelAsy Sub Assembly 2

Rashmi Pandit 2009-12-11 08:27:13

Rashmi,That works. Thank you!!!Thanks a lot for helping me.-George

2009-12-11 14:20:07

Rashmi,One more question...How can i call the second xsl file in the first file and provide the xml file generated from the fist as an input to it.I tried couple of things but i have not been successfulThanks for any help-George

2009-12-11 21:25:28

You can pass the second xslt as an input parameter to the first xslt. Let me know if you need code sample for that too.

Rashmi Pandit 2009-12-12 18:28:05

Rashmi, Wish you a great New year.could you please send me a code sample for 'pass the second xslt as an input parameter to the first xslt'Thanks again!

2010-01-04 21:59:15

Hi George, Happy New Year to you too! To pass an xslt through code you need to use the appropriate class. For e.g. in C# an instance of XsltArgumentList is passed to the XslCompiledTransform.Transform(). The xslt arg will be a name value pair. Consider name as 'driver' and the value as XmlDocument returned from the output of first xslt. In that case your xslt will change too. You should declare the parameter in the xslt: <xsl:param name="driver" /> and your xsl:for-each should use '$driver/root/Data[....'

Rashmi Pandit 2010-01-05 04:13:30

Hi Rashmi,Couple of questions...Is there a way where all this can be done without using any programming languagesomething along the lines where xslt2 is embedded into xslt1.xslt1 would write out the file to a certain location ex. c:\docs\output.xmlxslt2 is embedded in xslt1 and would take as an input c:\docs\output.xml.I'm trying to achieve a result where only one xslt will give me the required output. One constraint is that I dont have the option of writting C# etc. I have to get this done using an xslt.Any thoughts or help on this would be great.

2010-01-05 15:47:13

Yes, if you are not using any programming language and the output file for the first xslt is in a fixed location, you can define the driver in your xslt as: <xsl:variable name="driver" select="document('C:\XML\Output1.xml')"/>

Rashmi Pandit 2010-01-06 04:06:38

Thank you Rashmi.I have a new problem now. I can only call one xslt to format the document. Is there a way to call the second XSLT in the first one and pass the output xml file to it. Or preferably can it be done by creating a template within the first XSLT for the operations performed by the second xslt and call the template and pass it the out put xml file.please let me know what you think.-george

2010-02-16 16:59:56

Also, could you please provide code snippets. It would be great if this can be achieved by using templates.Thank you-george

2010-02-16 17:05:46

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