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I have some boost::property_tree::ptree. I need tree with removing some elements with certain tag name. For example, xml for source ptree is the following:

<?xml version="1.0" encoding="utf-8"?>
<document>
  <B atr="one" atr1="something">
    <to_remove attr="two">10</to_remove>
  </B>
  <to_remove>
    <C>value</C>
    <D>other value</D>
  </to_remove>
  <E>nothing</E>
</document>

And I'd like to get ptree with xml like the following:

<?xml version="1.0" encoding="utf-8"?>
<document>
  <B atr="one" atr1="something" />
  <E>nothing</E>
</document>

How to write function, that generate new ptree with removed <to_remove> nodes?

Loom
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2 Answers2

1

The value_type of ptree is std::pair< const Key, self_type >, so you can iterate the tree and remove corresponding nodes. The following is a sample.

void remove(ptree &pt){
using namespace boost::property_tree;
    for (auto p = pt.begin(); p != pt.end();){
        if (p->first == "to_remove"){
            p = pt.erase(p);
        }
        else{
            remove(p->second);
            ++p;
        }
    }
}
cqdjyy01234
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0

UPDATE Replaced my answer due to the comment: I'd recommend, instead, to use a proper XML library.

I believe Boost PropertyTree uses a modified RapidXml internally (but it's an implementation detail, so I'm not sure I would depend on it). Here's my take on it using PugiXML, which is a modern, header-only, non-validating XML library:

#include <pugixml.hpp>
#include <iostream>

int main()
{
    pugi::xml_document doc;
    doc.load_file("input.txt");

    for (auto& to_remove : doc.select_nodes("descendant-or-self::to_remove/.."))
        while (to_remove.node().remove_child("to_remove"));

    doc.save(std::cout);
}

Prints

<?xml version="1.0"?>
<document>
    <B atr="one" atr1="something" />
    <E>nothing</E>
</document>
sehe
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  • Thank you. However, this solution doesn't fit because it's hardly determined by specific xml structure. – Loom Jun 16 '14 at 16:04
  • @Loom oh well, I don't have patience to show you exactly how to roll with property tree, I suppose you could read the docs too. I've updated my answer to show you what I'd do instead – sehe Jun 16 '14 at 21:28