sorting nodes

scubin · #1 (**permalink**) July 6th, 2009, 05:04 PM

Hi all,

Please help me in figuring out how to code for this xml transformation
Thwe final outcome of the xml should contain all employee with highest salary from different TypeCde (Taking into Consideration all Employees nodes)
Please let me know if u have any questions

Below is the XML i have to transform:
<Company>

<Employees>
<employee >
<TypeCde>FN</TypeCde>
<salary>100000</salary>
</employee>
<employee>
<TypeCde>CL</TypeCde>
<salary>95000</salary>
</employee>
</Employees>
<Employees>
<employee>
<TypeCde>UH</TypeCde>
<salary>89000</salary>
</employee>
<employee>
<TypeCde>FN</TypeCde>
<salary>97000</salary>
</employee>
</Employees>
<Company>
Below is the xml i need after transformation

Thanks

mhkay · #2 (**permalink**) July 6th, 2009, 05:31 PM

First you have to group by typeCode - the way you do that depends greatly on whether it's XSLT 2.0 or 1.0. Then sort each group and take the first in descending order.

Code:

<xsl:for-each-group select="//employee" group-by="typeCode">
  <xsl:for-each select="current-group()">
     <xsl:sort select="xs:integer(salary)" order="descending"/>
     <xsl:if test="position() = 1">
        <xsl:copy-of select="."/>
     </xsl:if>
  </xsl:for-each>
</xsl:for-each-group>

scubin · #3 (**permalink**) July 6th, 2009, 05:53 PM

I want it for XSLT version 1.0 ,seems <xsl:for-each group> is for version 2.0

can i know how to group by using version 1.o

Thanks fro quick reply

mhkay · #4 (**permalink**) July 6th, 2009, 06:52 PM

Search for "Muenchian Grouping" in the index to your favourite XSLT textbook.

scubin · #5 (**permalink**) July 7th, 2009, 02:32 AM

Thank you Michael that helped a lot and i was able to get result what i want but there is one more thing that troubling me ,in the below XML if i want Employee Id to be outputted along with
Input XML:
<Company>
<Employees>
<Id>123</Id>
<employee >
<TypeCde>FN</TypeCde>
<salary>100000</salary>
</employee>
<employee>
<TypeCde>CL</TypeCde>
<salary>95000</salary>
</employee>
</Employees>
<Employees>
<Id>234</Id>
<employee>
<TypeCde>UH</TypeCde>
<salary>89000</salary>
</employee>
<employee>
<TypeCde>FN</TypeCde>
<salary>10000</salary>
</employee>
</Employees>
<Company>
Below is the my output should be

<Company>
<Employees>
<employee >
<TypeCde>FN</TypeCde>
<salary>100000</salary>
<Id>123</Id>
<Id>234</Id>
</employee>
<employee>
<TypeCde>CL</TypeCde>
<salary>95000</salary>
<Id>123</Id>
</employee>
<employee>
<TypeCde>UH</TypeCde>
<salary>89000</salary>
<Id>234</Id>
</employee>
</Employees>
</Comapny>

based on my muenchian method below is my code snippet
<xsl:for-each select="//employee[generate-id(.) = generate-id(key('k',
TypeCde)[1])]">
<xsl:for-each select="key('k', TypeCde)">
<xsl:sort select="salary" data-type="number" order="descending"/>
<xsl:if test="position() = 1">
<xsl:copy-of select="."/>
</xsl:if>
</xsl:for-each>
</xsl:for-each>
what can i do to get the <Id> of Employees in output xml?

Thnaks in advance

mhkay · #6 (**permalink**) July 7th, 2009, 04:40 AM

Just replace the copy-of with an apply-templates select=".", and write a template rule for Employee something like

Code:

<xsl:template match="employee">
  <xsl:copy>
    <xsl:copy-of select="*"/>
    <xsl:copy-of select="../Id"/>
  </xsl:copy>
</xsl:template>

scubin · #7 (**permalink**) July 7th, 2009, 08:49 AM

Thank you very much michael for your reply but what i want is I need even the Emplotees Ids those having same salary in input XML also to be printed.The above code simply gives the first one that is having highest salary.If there are 2 or 3 Employee Ids having same highest salary I need those Employee Ids as well. Something like below .Can you tell me how to accomplish this
<Company>
<Employees>
<employee >
<TypeCde>FN</TypeCde>
<salary>1000</salary>
<Id>123</Id>
<Id>234</Id>
</employee>
<employee>
<TypeCde>CL</TypeCde>
<salary>95000</salary>
<Id>123</Id>
</employee>
<employee>
<TypeCde>UH</TypeCde>
<salary>89000</salary>
<Id>234</Id>
</employee>
</Employees>
</Comapny>

Thans a lot for your support

mhkay · #8 (**permalink**) July 7th, 2009, 08:57 AM

Then you need to do a second level of grouping (group by salary) before you sort these groups on salary and choose the first; you then print the Ids for all the employees in the group.

scubin · #9 (**permalink**) July 7th, 2009, 12:11 PM

Iam sorry michael ,I couldnt figure out how to do that second level grouping

on salary,Iam getting lot of duplicates if iam doing 2 level grouping.Can you

help on this please

Thanks

mrame · #10 (**permalink**) July 8th, 2009, 02:19 AM

Try this:

Code:

<xsl:key name="k" match="employee" use="TypeCde"/>

<xsl:template match="Company">
<xsl:copy>
<Employees>
<xsl:for-each select="//employee[generate-id(.) = generate-id(key('k',TypeCde)[1])]">
<xsl:for-each select="key('k', TypeCde)">
<xsl:sort select="salary" data-type="number" order="descending"/>
<xsl:if test="position() = 1">
<xsl:copy>
<xsl:variable name="type" select="TypeCde"></xsl:variable>
<xsl:variable name="sal" select="salary"></xsl:variable>
<xsl:copy-of select="$type"/>
<xsl:copy-of select="$sal"/>
<xsl:copy-of select="../Id"/>
<xsl:copy-of select="../..//Id[..//TypeCde = $type][not(..//salary = $sal)]"/>
</xsl:copy>
</xsl:if>
</xsl:for-each>
</xsl:for-each>
</Employees>
</xsl:copy>
</xsl:template>

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