XPath - Get node with no child of specific type

user_42230 0
Viewed 74628

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>
xpath
5 Answers

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" />
user_364980

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.

user_175160

If you are trying to get A anywhere in the hierarchy from the root, this works (for xslt 1.0 as well as 2.0 in case its used in xslt)

//descendant-or-self::node()[local-name(.) = 'a' and not(count(b))]

OR you can also do

//descendant-or-self::node()[local-name(.) = 'a' and not(b)]

OR also

//descendant-or-self::node()[local-name(.) = 'a' and not(child::b)]

There are n no of ways in xslt to achieve the same thing.

Note: XPaths are case-sensitive, so if your node names are different (which I am sure, no one is gonna use A, B), then please make sure the case matches.

user_392890
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