XPath operator "!=". How does it work?

Question

XML document:

<doc>     <A>            <Node>Hello!</Node>        </A>       <B>              <Node/>     </B>        <C>     </C>      <D/> </doc> 

How would you evaluate the following XPath queries?

/doc/A/Node != 'abcd'   /doc/B/Node != 'abcd'   /doc/C/Node != 'abcd'   /doc/D/Node != 'abcd'   

I would expect ALL of these to evaluate to true.

However, here are the results:

/doc/A/Node != 'abcd'     true /doc/B/Node != 'abcd'     true /doc/C/Node != 'abcd'     false /doc/D/Node != 'abcd'     false 

Is this expected behavior? Or is it a bug with my XPath provider (jaxen)?

Answer 1

Recommendation: Never use the != operator to compare inequality where one or both arguments are node-sets.

By definition the expression:

$node-set != $value 

evaluates to true() exactly when there is at least one node in $node-set such that its string value is not equal to the string value of $value.

Using this definition:

$empty-nodeset != $value  

is always false(), because there isn't even a single node in $empty-nodeset for which the inequality holds.

Solution:

Use:

not($node-set = $value) 

Then you get all results true(), as wanted.