Need help on xsl/xquery desired ouput

Question

My requirement is like below:

Input :-

<Root>
<Name>abc</Name>
<Name>def</Name>
<Name>abc</Name>
<Name>abc</Name>
<Name>def</Name>
<Name>def</Name>
</Root>

Output:- I need to make xml of Name element who's value don't be repititive

<Root>
<Name>abc</Name>
<Name>def</Name>  
</Root>

We can use xsl or xquery anything... Please help me on the same...

http://stackoverflow.com/questions/2291567/how-to-use-xslt-to-create-distinct-values — Magicianred, Jul 27 '14 at 05:02
Hi, It worked when there is no namespace defined with xml. But In real case, xml will contains namespace like below:- abc def abc abc def def — user3384223, Jul 27 '14 at 09:46
hey tried with localname convention. It worked.. thanks alot for your super quick help... — user3384223, Jul 27 '14 at 09:55
Thanks to @Mads Hansen in http://stackoverflow.com/questions/2291567/how-to-use-xslt-to-create-distinct-values Enjoy your code :) — Magicianred, Jul 27 '14 at 09:58

score 2 · Accepted Answer · answered Jul 27 '14 at 07:01

2

In XSLT 2.0, use

<xsl:for-each-group select="/*/Name" group-by=".">
  <xsl:copy-of select="current-group()[1]"/>
</xsl:for-each>

In XQuery 1.0 use

for $n in distinct-values(/*/Name)
return <Name>{$n}</Name>

answered Jul 27 '14 at 07:01

Michael Kay

score 0 · Answer 2 · answered Jul 29 '14 at 08:33

0

In addition to @michael-kay's excellent answer, if you have XQuery 3.0 then you could use:

distinct-values(/*/Name) ! <Name>{.}</Name>

The ! is the Simple Map Operator, which IMHO can be nicer for such simple one-line transformation cases than a for.

answered Jul 29 '14 at 08:33

adamretter

2 Answers2