Getting the parent node attributes in XSL

Question

In my XML I have the following:

<a>
  <b>
    <c something="false">
      <d>
        <e>
          <f>someResult</f>
        </e>
      </d>
    </c>
  </b>
</a>

Now in the XSL within a loop I can do the following:

<xsl:value-of select="f"></xsl:value-of>

But how can I get the attribute in c?

I've tried doing the following

<xsl:value-of select="////@something"></xsl:value-of>

As well as trying parent and nothing seems to be working. Can you get parent nodes like this?

Also, I cannot just do:

<xsl:value-of select="/a/b/c/@something"></xsl:value-of>

As there can be multiple of c.

score 51 · Accepted Answer · edited May 01 '13 at 17:46

51

To move up the tree you use ".." per level ie in this instance probably

select="../../../@something"

You can also select an ancestor node by name (approx)

select="ancestor::c[1]/@something"

edited May 01 '13 at 17:46

demongolem

answered Feb 28 '12 at 16:25

kaj

score 12 · Answer 2 · answered Feb 28 '12 at 17:08

12

Use:

ancestor::c[1]/@something

This selects the attribute named something of the first (from the current node upwards) ancestor named c.

answered Feb 28 '12 at 17:08

Dimitre Novatchev

1

@AkashTantri, no, `ancestor::c` is the node-list of *all* ancestor nodes that are named "c". `ancestor::c[1]` selects the nearest "c" ancestor of the current node. – Dimitre Novatchev Apr 18 '18 at 15:56

2 Answers2