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Answer 1

+5 A:

In XPath 1.0:

/Message/Customers/Customer[not(@pin=/Message/Addresses/Address/@pin1)]

Alejandro 2010-08-14 17:51:43

Wow it works well. Thank you so much.

SaravananArumugam 2010-08-14 18:00:40

@Saravanandss: You are wellcome!

Alejandro 2010-08-14 18:15:18

+1. This is fundamental XPath knowledge -- unfortunately the OP is happy that "it works" and would probably not delve further to understand what actually is happening... :(

Dimitre Novatchev 2010-08-14 18:53:20

I think the fundamental what-is-happening that @Dimitre refers to is the fact that the "=" operator is an "existentially quantified comparison" operator, that is, `A = B` tells you whether there is a pair of values, one in nodeset A and one in nodeset B, that are equal to each other. In the above example, A is @pin which addresses only one node (because it is in the context of just one `Customer`), while the B path addresses many nodes.

LarsH 2010-08-18 18:34:07

Answer 2

+2 A:

An alternative to the good answer by @Alejandro, which I upvoted, is the following transformation, which uses keys and will be more efficient if the number of existing customers is big:

<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:output omit-xml-declaration="yes" indent="yes"/>

    <xsl:key name="kexistingByPin"
         match="Address" use="@pin1"/>

 <xsl:template match="node()|@*">
     <xsl:copy>
       <xsl:apply-templates select="node()|@*"/>
     </xsl:copy>
 </xsl:template>

 <xsl:template match="/">
   <xsl:apply-templates select=
    "*/*/Customer[not(key('kexistingByPin', @pin))]"/>
 </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the provided XML document:

<Message>
    <Customers>
        <Customer pin="06067">1</Customer>
        <Customer pin="06068">2</Customer>
        <Customer pin="06069">3</Customer>
        <Customer pin="06070">4</Customer>
        <Customer pin="06072">5</Customer>
    </Customers>
    <Addresses>
        <Address pin1="06067">A</Address>
        <Address pin1="06068">B</Address>
        <Address pin1="06069">C</Address>
    </Addresses>
</Message>

the wanted, correct answer is produced:

<Customer pin="06070">4</Customer>
<Customer pin="06072">5</Customer>

Dimitre Novatchev 2010-08-14 19:07:38

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Not IN equivalent in XPath expression

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