XSLT, XPath unique child nodes only problem where non unique nodes are not selected at all -

 <root>   <parent>     <child>      <name>john</name>     </child>     <child>       <name>ben</name>     </child>   </parent>   <parent>     <child>      <name>john</name>     </child>     <child>      <name>mark</name>     </child>     <child>       <name>luke</name>     </child>  </parent> </root> 

i want unique child nodes i.e. 1 child node if there more 1 same name.

such as:

john ben mark luke 

i have tried:

 <xsl:for-each select="parent">   <xsl:for-each select="child[name != preceding::name]">     <xsl:value-of select="name"/>   </xsl:for-each> </xsl:for-each> 

but get:

ben mark luke 

?!

your problem using != operator comparison between value , node-set.

this wrong -- avoid using != operator , use not() function , = operator when 1 of operands in comparison node-set.

below correct solution:

<xsl:stylesheet version="1.0"  xmlns:xsl="http://www.w3.org/1999/xsl/transform">  <xsl:output omit-xml-declaration="yes" indent="yes"/>  <xsl:strip-space elements="*"/>   <xsl:template match="/*">     <xsl:for-each select="parent">       <xsl:for-each select="child[not(name = preceding::name)]">         <xsl:value-of select="concat(name, ' ')"/>       </xsl:for-each>     </xsl:for-each>  </xsl:template> </xsl:stylesheet> 

when transformation applied on provided xml document:

<root>   <parent>     <child>      <name>john</name>     </child>     <child>       <name>ben</name>     </child>   </parent>   <parent>     <child>      <name>john</name>     </child>     <child>      <name>mark</name>     </child>     <child>       <name>luke</name>     </child>  </parent> </root> 

the wanted, correct result produced:

john ben mark luke  

explanation: here how w3c xpath 1.0 spec defines semantics of != operator:

"if 1 object compared node-set , other string, comparison true if , if there node in node-set such result of performing comparison on string-value of node , other string true."

this means

's' != node-set 

is true if there 1 node in node-set isn't equal 's'.

this isn't semantics wanted.

on other side,

not('s' = node-set())  

is true only if there isn't node in node-set equal 's'.

this wanted comparison.

do note: grouping technique have chosen o(n^2) , should used on small sets of values dedupped. if efficiency needed, means use muenchian method grouping (discussing or demo-ing falls outside scope of question).