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//p[not(ancestor::*[3])]
  //table[ancestor::*[1][self::p] or ancestor::*[2][self::p]]
    tr/td//a[ancestor::*[1`][self::td] or ancestor::*[2][self::td]]
heyjude
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3 Answers3

3
//                               # from the root node, look at all descendants
p[                               # select nodes of type <p>, who have…
  not(ancestor::*[3])            #   …no ancestor 3 levels up
]                                #
//                               # from these nodes, select descendants
table[                           # of type <table>, who have…
  ancestor::*[1][self::p]        #   …a <p> as their direct ancestor
  or                             #   or
  ancestor::*[2][self::p]        #   …a <p> as their second ancestor
]                                # 
                                 # syntax error, this should be a location step
tr                               # …select all nodes of type <tr>
/                                # from their children…
td                               # …select all nodes of type <td>
//                               # from their descendants…
a[                               # …select all nodes of type <a>, who have
  ancestor::*[1][self::td]       #   …a <td> as their direct ancestor   
  or                             #   or 
  ancestor::*[2][self::td]       #   …a <td> as their second ancestor
]

Or, expressed in HTML:

<html>
  <body>
    <p>
      <table>
        <tr> 
          <td>
            <a title="These would be selected." />
          </td>
        </tr>
      </table>
    </p>
  </body>
</html>

The whole XPath does not make too much sense anyway. It goes without saying that <p><table> is invalid HTML.

Tomalak
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2

Lets break this down:

//p[not(ancestor::*[3])]

Selects all p tags who do not have a 3rd ancestor.

In those: 

//table[ancestor::*[1][self::p] or ancestor::*[2][self::p]]

It selects all table tags whose first or second ancestor is a p tag.

Then:

tr/td//a[ancestor::*[1`][self::td] or ancestor::*[2][self::td]]

This isn't entirely correct (there should be a / at the start). However, it goes down the tr/td//s to select all a tags whose first or second ancestor is a td tag.

All and all, it is very convoluted and could probably be a lot easier to achieve with some id attributes defined in the relevant places.

Oded
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    Having written a fair number of xpaths for the purpose of screen scraping, I'd say this one is horribly broken. Not only was it not well-formed, but it's incredibly fragile. It'd break at the slightest change to the page. – Brian Donovan Dec 31 '10 at 16:52
1

The XPath expression you specified:

//p[not(ancestor::*[3])]
     //table[ancestor::*[1][self::p] or ancestor::*[2][self::p]]
           tr/td//a

isn't syntactically legal -- it is missing a / before the tr.

The corrected XPath expression:

//p[not(ancestor::*[3])]
     //table[ancestor::*[1][self::p] or ancestor::*[2][self::p]]
           /tr/td//a

was provided as answer to this question.

As explained in the linked (above) answer, the meaning is:

This selects all a elements whose parent or grand-parent is td, whose parent is a tr, whose parent is a table, whose parent or grandparent is a p that has less than 3 ancesstor - elements

The OP wanted a way to get the a elements that can be located under a p buried not deeper than 3 levels beneath the root of the document, then under a table/tr/td where the table is buried at a level not greater than 3 from the p.

Certainly, wanting to select such nodes may not seem too-meaningful, but we are in no position to judge anyone's needs and requirements.

The amazing fact is that XPath is so powerful to satisfy even such requirements.

Community
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Dimitre Novatchev
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  • A minor: Isn't `ancestor::*[1][self::p]` equal to `parent::p`? It looks like you'd used that to make the expression readable... –  Jan 01 '11 at 21:44
  • @Alejandro: Exactly, this is for the user to see the pattern (`ancesstor::*[N][self::p]`), so I think uniformness is helpful here. BTW, Happy New Year! :) – Dimitre Novatchev Jan 01 '11 at 22:11
  • Thanks! And happy New Year to you, too! –  Jan 01 '11 at 22:16