Car*_*bou 8 c++ boost boost-propertytree
我有一个XML结构:
<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>
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这是被读入boost::property_tree,有1..Many <SomeElement>秒,然后在该单元内任意深度有可能是1..Many <ElementIWant>小号
有没有办法<ElementIWant>按照它们在doc中出现的顺序直接(在单个循环中)迭代?
我看过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;
}
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但是它似乎无法返回节点(我怀疑是因为equal_range在提供的树节点级别工作)这让我想到上面的问题......
不可能直接迭代所有元素; 文件说
没有办法迭代整个树.
现在,您可以使用递归,并在每个级别应用STL算法来模拟它; 在下面的示例中,它不符合您在单个循环中执行此操作的要求,但它确实有效:
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;
}
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这是一个"完全"递归的版本并保留外观的顺序:
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);
}
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