XSLT split output files - muenchian grouping

Question

I have an XSLT file so as to transform large amount of data. I would like to add a "split" functionality, either as a chained XSLT or within the current XSLT that can create multiple output files so as to limit the size of the files under a certain threshold. Let's assume that the input XML is as below:

<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>
<Person>             
<name>Jack</name>             
<date>June19</date>             
<workTime taskID="1">20</workTime>             
<workTime taskID="2">30</workTime>             
</Person>    
</People>

The XSLT file is as below using muenchian grouping.

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

However, I'd like ,as an output, several files as below instead of a large one. For this example, I have set only one name per file..but this should be a parameter.

Output file for first person:

<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>

Output file for second person:

<People>
    <Person>
        <name>Jack</name>
        <taskID>1</taskID>
        <workTime>
        <date>June19</date>
        <time>20</time>
        </workTime>
    </Person>
    <Person>
        <name>Jack</name>
        <taskID>2</taskID>
        <workTime>
        <date>June19</date>
        <time>30</time>
        </workTime>
    </Person>
</People>

What would be the preferred and most elegant approach using XSLT 1.0? Is there a way to call a chained XSLT within the XSLT so as to split the output files?

Cheers.

Good question, +1. See my answer for a number of ways to do this. — Dimitre Novatchev, Nov 12 '10 at 23:59

Dimitre Novatchev · Accepted Answer · 2010-11-13T02:38:36.063

6

Is there a way to call a chained XSLT within the XSLT so as to split the output files?

A few ways:

You could write an extension function to do this -- check the documentation of your XSLT processor.
Use the <exsl:document> extension element of EXSLT, in case this is supported by your XSLT processor
Use the <saxon:output> extension element if you have Saxon 6.x
In a loop from your programming language invoke a separate transformation, passing to it as parameter the name of the person for which to produce results.

Here are code examples for 2. and 3. above:

Using <saxon:output> :

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:saxon="http://icl.com/saxon"
  extension-element-prefixes="saxon" >

 <xsl:template match="/">
  <xsl:for-each select="/*/*[not(. > 3)]">
   <saxon:output href="c:\xml\doc{.}">
    <xsl:copy-of select="."/>
   </saxon:output>
  </xsl:for-each>
 </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the following XML document:

<nums>
  <num>01</num>
  <num>02</num>
  <num>03</num>
  <num>04</num>
  <num>05</num>
  <num>06</num>
  <num>07</num>
  <num>08</num>
  <num>09</num>
  <num>10</num>
</nums>

three files: c:\xml\doc1 , c:\xml\doc2 and c:\xml\doc3 are created with the wanted contents.

The same example using <exslt:document>:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:ext="http://exslt.org/common"
  extension-element-prefixes="saxon" >

 <xsl:template match="/">
  <xsl:for-each select="/*/*[not(. > 3)]">
   <ext:document href="c:\xml\doc{.}">
    <xsl:copy-of select="."/>
   </ext:document>
  </xsl:for-each>
 </xsl:template>
</xsl:stylesheet>

edited Nov 13 '10 at 02:38

answered Nov 12 '10 at 23:59

Dimitre Novatchev

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I think exsl:document seems to be the best idea. Do you have any insight on how it would work for the particular example ? – Daniel Nov 13 '10 at 00:43
@Daniel: I added two code examples: for `` and for `` – Dimitre Novatchev Nov 13 '10 at 02:39
@Dimitre. Thanks and do you have some insight as to how your solution can be mized with the grouping XSLT in the question? – Daniel Nov 13 '10 at 14:55
@Daniel: Please, do your homework. Read about Muenchian Grouping. In the grouping you come up with just put the `` or `` inside the `` that selects all unique nodes. Your question is about how to write multiple document results -- not how to do grouping. In case you have problems with grouping, just read (a lot of good questions and answers are at SO) or ask a separate question. – Dimitre Novatchev Nov 13 '10 at 16:36
@Daniel: If you can use EXSLT, that means that you aren't ussing MSXSL. So, you also could update your XSLT processor to version 2.0 – Nov 14 '10 at 20:42
@DimitreNovatchev Do you know if the `exsl:document` solution can be run somehow with the XSL processor included in the Internet Explorer 8.0? Because of many reasons and the tools that are available at my current workplace I have to run all my XSL transformations via this version of IE and unfortunately I keep getting an error `Element ext:document is not known extension element`. And one more remark, should not there be `extension-element-prefixes = ext` instead of `extension-element-prefixes = saxon`in the aforementioned example? – Jagger Jan 22 '14 at 13:50
@Jagger, No, IE only uses MSXML (3 and 6 -- depending the version of IE), and MSXML doesn't implement any EXSLT extensions. – Dimitre Novatchev Jan 22 '14 at 15:14
@Jagger, But you could use SaxonCE, which implements XSLT 2.0 and that means, among other things, – Dimitre Novatchev Jan 22 '14 at 18:27
@DimitreNovatchev Trust me, if I could I would. :-) On my private computer I have been using Saxon and XSLT 2.0 for a long time. However currently I am working for a big bank, where there is a set of standard software and requesting a thing that is not standard would cost too much time as the installation of such software would first generate a lot of questions, second many people would have to agree on that. The transformation will end up in the end in an SAP system, where only XSLT 1.0 processor is available. It might be interesting if it supports `exsl`. I will let you know. – Jagger Jan 23 '14 at 07:40

XSLT split output files - muenchian grouping

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