XPath - Get node with no child of specific type

Question

XML: /A/B or /A

I want to get all A nodes that do not have any B children.

I've tried

/A[not(B)]  
/A[not(exists(B))]

without success

I prefer a solution with the syntax /*[local-name()="A" and .... ], if possible. Any ideas that works?

Clarification. The xml looks like:

<WhatEver>
  <A>
    <B></B>
  </A>
</WhatEver> 

or

<WhatEver>
  <A></A>
</WhatEver>

Answer 1

Maybe *[local-name() = 'A' and not(descendant::*[local-name() = 'B'])]?

Also, there should be only one root element, so for /A[...] you're either getting all your XML back or none. Maybe //A[not(B)] or /*/A[not(B)]?

I don't really understand why /A[not(B)] doesn't work for you.

~/xml% xmllint ab.xml
<?xml version="1.0"?>
<root>
    <A id="1">
            <B/>
    </A>
    <A id="2">
    </A>
    <A id="3">
            <B/>
            <B/>
    </A>
    <A id="4"/>
</root>
~/xml% xpath ab.xml '/root/A[not(B)]'
Found 2 nodes:
-- NODE --
<A id="2">
    </A>
-- NODE --
<A id="4" />

Answer 2

Try this "/A[not(.//B)]" or this "/A[not(./B)]".

Answer 3

The first / causes XPath to start at the root of the document, I doubt that is what you intended.

Perhaps you meant //A[not(B)] which would find all A nodes in the document at any level that do not have a direct B child.

Or perhaps you are already at a node that contains A nodes in which case you just want A[not(B)] as the XPath.