How to iterate over XML structure in boost::property_tree

Question

I have an XML structure along the lines of:

<root>
 <SomeElement>
  <AnotherElement>
   <ElementIWant x="1" y="1"/>
  </AnotherElement>
 </SomeElement>
 <SomeElement>
  <AnotherElement>
   <ElementIWant x="1" y="1"/>
   <ElementIWant x="2" y="1"/>
   <ElementIWant x="3" y="1"/>
  </AnotherElement>
 </SomeElement>
</root>

Which is being read into a boost::property_tree, There are 1..Many <SomeElement>s, and then at an arbitrary depth within that element there could be 1..Many <ElementIWant>s

Is there a way to iterate over the <ElementIWant> directly (in a single loop) in the order that they appear in the doc?

I have looked at equal_range

void iterateOverPoints()
{
     const char* test = 
     "<?xml version=\"1.0\" encoding=\"utf-8\"?><root>"
      "<SomeElement>"
       "<AnotherElement>"
        "<ElementIWant x=\"1\" y=\"1\"/>"
       "</AnotherElement>"
      "</SomeElement>"
      "<SomeElement>"
       "<AnotherElement>"
        "<ElementIWant x=\"1\" y=\"1\"/>"
        "<ElementIWant x=\"2\" y=\"1\"/>"
        "<ElementIWant x=\"3\" y=\"1\"/>"
       "</AnotherElement>"
      "</SomeElement>"
    "</root>";

    boost::property_tree::ptree message;
    std::istringstream toParse(test); 
    boost::property_tree::read_xml(toParse,result_tree);

    //Now we need to locate the point elements and set the x/y accordingly.
    std::pair< boost::property_tree::ptree::const_assoc_iterator,
               boost::property_tree::ptree::const_assoc_iterator > result =
         message.equal_range("ElementIWant");

    for( boost::property_tree::ptree::const_assoc_iterator it = result.first; 
           it != result.second; ++it )
    {
        std::cout  << it->first << " : ";
        const boost::property_tree::ptree& x = it->second.get_child( "<xmlattr>.x" );
        const boost::property_tree::ptree& y = it->second.get_child( "<xmlattr>.y" );
        std::cout << x.get_value<int>() << "," << y.get_value<int>() << "\n";
    }

    return;
}

However it seems to fail to return nodes (Which I suspect is because equal_range works at the level of the tree node supplied) Which brings me to the question above...

Belmonte

Did you try equal_range("SomeElement.AnotherElement.ElementIWant");? Not sure what that will do when there are two copies of SomeElement, though. – Podagra

Blast from the past! @TreborRude I think at the time it didn't behave in the way I wanted, or it didn't work. Perhaps the library has been enhanced since to do it - but I'm well away from that area of code at the moment. :) Thanks for the suggestion though. – Belmonte

Answer 1

It is not possible to iterate over all elements directly; the documentation says

There is no way to iterate over the entire tree.

Now, you could use recursion, and apply STL algorithms at each level to mimic that; it does not fit your requirement of doing this in a single loop in my sample below, but it does works:

template <typename InputIt, typename OutputIt, typename Compare>
void collect(InputIt first, InputIt last, OutputIt dest, Compare comp)
{
    typedef typename std::iterator_traits<InputIt>::reference reference;

    std::copy_if (
        first, last, dest,
        [comp] (reference what) { return comp(what.first); });

    std::for_each (
        first, last,
        [dest, comp] (reference what) { collect(what.second.begin(), what.second.end(), dest, comp); });
}


std::vector<std::pair<std::string, ptree>> match;

collect(
    xml.begin (), xml.end (), std::back_inserter(match),
    [] (const std::string& key) { return key == "ElementIWant"; });

for (auto pair: match)
{
     std::cout << pair.first << std::endl;
}

---

Here is a version that is "fully" recursive and preserve the order of appearance:

template <typename InputIt, typename OutputIt, typename Compare>
void collect_recursive(InputIt first, InputIt last, OutputIt dest, Compare comp)
{
    typedef typename std::iterator_traits<InputIt>::reference reference;

    if (first == last)
    {
        return;
    }

    auto begin = first->second.begin ();
    auto end = first->second.end ();

    if (begin != end)
    {
        collect_recursive (begin, end, dest, comp);
    }

    if (comp (first->first))
    {
        dest = *first;
    }

    collect_recursive (++first, last, dest, comp);
}

Jeffryjeffy

I was hopeful that there might be some trick, but I'm playing with something similar now as well. I'll see if I can incorporate what you've done above - Thx for the reply. – Belmonte

BTW will accept this as it is the right answer I suspect but want to play for a while before doing so :) – Belmonte

@Belmonte Actually, I am playing a bit on my side as well, I'll edit with a "fully" recursive version, because in this first attempt, the order of appearance is not conserved! – Jeffryjeffy