Getting rid of Duplicate nodes

sureshvadali · July 31st, 2012, 06:05 AM

Hello Everyone,

I have pretty similar looking problem which have got many solutions over the web but none of them have worked well for me so far. Here's the problem statement:

XML file version "1.0":
<enums>
<enum name="enumOne">
<val v="1"/>
</enum>
<enum name="enumTwo">
<val v="2"/>
</enum>
<enum name="enumThree">
<val v="3"/>
</enum>

<table>
<child name="child1">
<type t="enumOne"/>
..........
.........
</child>
<child name="child2">
<type t="enumTwo"/>
..........
.........
</child>
<child name="child3">
<type t="enumOne"/>
..........
.........
</child>
</table>

XSL version "1.0": It would traverse through child node and retrieve the enum type i.e. enumXXX from <type> tag. and find it's definition from <enums>

<xsl:for-each select="/child/type">
<xsl:call-template name="getenumdefs">
<xsl:with-param name="enumname">
<xsl:value-of select="@t"/>
</xsl:with-param>
</xsl:call-template>
<xsl:for-each>

<xsl:template name="getenumdefs">
<xsl:param name="enumname"/>
<xsl:for-each select="/enums/enum">
<xsl:if test="$enumname = '$name'">
<xsl:value-of select="val/@v"/>
</xsl:if>
</xsl:for-each>

Problem with this approach is, it would print the duplicate entries for "enumOne" as both child1 and child3 uses it. There are multiple solutions available in the form of preceding-sibling or checking generate-id(), but nothing had really worked well for this xml/xsl. Appreciate if anyone can provide any inputs/suggestions.

mhkay · July 31st, 2012, 06:32 AM

Are you constrained to use XSLT 1.0? Are you familiar with Muenchian grouping?

sureshvadali · July 31st, 2012, 07:03 AM

The other alternative could be: create a flag variable for the first occurrence of enum and flag it. The subsequent calls to the template getenumdefs would check if the flag is set for the respective enum and skip it if flagged. But am not sure about the feasibility of such implementation in XSL.

mhkay · July 31st, 2012, 07:25 AM

Flag variables? No, you are thinking much too procedurally.

Why didn't you answer my questions?

sureshvadali · August 1st, 2012, 09:06 AM

Sorry Michael, I replied to my original thread instantly after posting it, but it seems there was a delay in capturing it into the thread. I just saw your responses, thanks for that. Right most of XSL's have been written in 1.0 and the source XML also adheres to 1.0 version moving to 2.0 might be tricky?

Though am not familiar about Muenchian grouping but I went through some examples and figured out the below approach:

<xsl:key name="attrenum" match="type" use="@t" />
<xsl:template match="/root/parent">
<xsl:for-each select="table">
<xsl:variable name="tab">
<xsl:value-of select="@name"/>
</xsl:variable>
<xsl:if test="$tab= 'two'"> <-- I added a name to the table with "one" "two" ..
<xsl:variable name="attribute">
<xsl:value-of select="@name"/>
</xsl:variable>
<xsl:for-each select="child">
<xsl:for-each select="type[count(. | key('attrenum', @t)[1]) = 1]">
<xsl:value-of select="@t" /> <br/>
</xsl:for-each>
</xsl:for-each>
</xsl:if>
</xsl:for-each>
</xsl:template>

But it fails if the table being searched is deep.

mhkay · August 1st, 2012, 09:32 AM

If your only reason for not moving to XSLT 2.0 is that you think it might be tricky, then make the move. Any time it takes will be repaid amply within a couple of weeks.

sureshvadali · August 2nd, 2012, 09:06 AM

I gave some thoughts to the challenges arriving out of XSLT 1.0 for this problem and eventually made up my mind to move on to XSLT 2.0. With some web search I found out that xsl:for-each-group or distinct-values() could be an alternative, do you have any other better options to suggest?

for-each-group:

<xsl:for-each-group select="//type" group-by="@t">
<xsl:copy-of select="current-group( )[1]"/>
</xsl:for-each-group>

distinct-values():

<xsl:for-each select="distinct-values(type/@t)">
<xsl:sort/>
<xsl:value-of select="."/> <xsl:call-template name="newline"/>
</xsl:for-each>

But I think both are flawed, anything needs to be corrected in the above statements?

Thanks

Martin Honnen · August 2nd, 2012, 09:13 AM

As far as I can see you havenÂ´t really explained what result you want. Your attempt with for-each-group looks fine to group "type" elements by the "t" attribute value and to extract the first item in each group, although doing

Code:

<xsl:copy-of select="."/>

suffices instead of the current-group()[1] you have.

sureshvadali · August 2nd, 2012, 02:26 PM

Thanks Martin for the response, my intention is to identify first occurrence of @t attribute in the group, but it has given an empty output, the XML which I mentioned out here is little simple and is just a prototype. The XSL used here is applied on a complex XML file which has got ~250 "type" elements something similar to:

Code:

<root>
  <parent>
	<table>
	<attr ......>
	   <type t="enumone">
	   <info>
		<name .....>
	   </info>
	</attr>

	<attr>
	    <type t="int">
		<range min="1" max="255"/>
	     </type>
	     <info>
		<name .....>
	   </info>
	</attr>

	<attr>
	    <type t="string">
	     <info>
		<name .....>
	   </info>
	</attr> 
	<attr ......>
	   <type t="enumtwo">
	   <info>
		<name .....>
	   </info>
	</attr>

	<attr>
	    <type t="float">
		<range min="1.0" max="25.5"/>
	     </type>
	     <info>
		<name .....>
	   </info>
	</attr>

	<attr>
	    <type t="int">
	     <info>
		<name .....>
	   </info>
	</attr> 


	<attr>
	    <type t="enumone">
	     <info>
		<name .....>
	   </info>
	</attr> 
	<attr>
	    <type t="enumthree">
	     <info>
		<name .....>
	   </info>
	</attr> 
	<attr>
	    <type t="enumone">
	     <info>
		<name .....>
	   </info>
	</attr> 
    </parent>
 </root>

sureshvadali · August 3rd, 2012, 04:59 AM

The expected output is:

enumone
int
string
enumtwo
float
enumthree