Re: [netmod] default namespace in XPath

[wangzitao]

Hi Lada,

In my understanding, it means to equal to current module's prefix. 
For example: 

module ietf-foo {
     namespace "urn:ietf:params:xml:ns:yang:ietf-foo";
     prefix foo;
        
        import ietf-interfaces {
    prefix if;
    }
        
    .....

         augment "/if:interfaces/if:interface"{
          list A{
           key "a";
           leaf a{
            type string;
           }
          }
         }
         
         list B{
          key "b";
          leaf b{
           type leafref{
            path "/if:interfaces/if:interface/A/a";
           }
          }
         }
         
}

In above example, the path "/if:interfaces/if:interface/A/a" equal to 
"/if:interfaces/if:interface/foo:A/foo:a".

BR!
-Michael

-----邮件原件-----
发件人: netmod [mailto:netmod-boun...@ietf.org] 代表 Ladislav Lhotka
发送时间: 2016年11月24日 17:14
收件人: NETMOD WG
主题: [netmod] default namespace in XPath

Hi,

second bullet in sec. 6.4.1 of RFC 7950 says:

    Names without a namespace prefix belong to the same namespace as the 
identifier of the current node.

But what is the current node? Is it the same as the result of the current() 
function or, if we have

    "/x:foo/bar"

is the namespace of "bar" the one corresponding to prefix "x"?

Lada

--
Ladislav Lhotka, CZ.NIC Labs
PGP Key ID: E74E8C0C




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