我正在尝试使用Boost C ++的ptree解析以下SVG(XML)标记......
<?xml version="1.0" standalone="no"?>
<!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 1.1//EN" "http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd">
<svg height="395.275590551181" version="1.1"
viewBox="0 0 757.48031496063 395.275590551181" xmlns="http://www.w3.org/2000/svg"
xmlns:xlink="http://www.w3.org/1999/xlink" style="background-color: #ffffff;">
<desc>Some Description</desc>
<g visibility="hidden" pointer-events="all">
<rect x="0" y="0" width="100%" height="100%" fill="none" />
</g>
<g>
<path d="M 5 56 L 5 14 L 8 11 L 34 11 Z" />
<path d="M 10 100 L 10 140 L 89 131 L 324 2 Z" />
</g>
<g>
<path d="M 20 130 L 1 1 L 89 130 L 34 2 Z" />
</g>
</svg>
typedef boost::property_tree::ptree::value_type vt;
boost::property_tree::ptree root;
read_xml("My_SVG.svg", root);
BOOST_FOREACH (vt const &nodes, root.get_child("svg")) {
//only action on g's and not desc, comments, etc.
if(nodes.first=="g"){
boost::property_tree::ptree g = nodes.second;
//only action on g's that contain paths, not g->rect, for example.
if(g.count("path") != 0){
BOOST_FOREACH (vt const &p, g.get_child("path")) {
std::cout << p.second.get("d", "false") << std::endl;
}
}
}
}
M 5 56 L 5 14 L 8 11 L 34 11 Z
M 20 130 L 1 1 L 89 130 L 34 2 Z
它编译得很好,但它没有拿起M 10 100 L 10 140 L 89 131 L 324 2 Z
条目。为什么第二个BOOST_FOREACH
没有逐步通过每个<path>
和cout
。
答案 0 :(得分:3)
第二个BOOST_FOREACH正在调用g.get_child("path")
,它返回名为g
的{{1}}的第一个孩子。您需要使用您要查找的密钥迭代所有子项。
这可以通过以下方式实现:
path
e.g。
/** Find the range of children that have the given key. */
std::pair<assoc_iterator, assoc_iterator>
ptree::equal_range(const key_type &key);
/** Count the number of direct children with the given key. */
size_type count(const key_type &key) const;
我发现以下代码在使用ptree时很有用,它会显示整个解析后的ptree:
typedef boost::property_tree::ptree::value_type vt;
BOOST_FOREACH (vt const &nodes, pt.get_child("svg"))
{
//only action on g's and not desc, comments, etc.
if(nodes.first=="g")
{
const boost::property_tree::ptree& g = nodes.second;
//only action on g's that contain paths, not g->rect, for example.
if(g.count("path") != 0)
{
std::pair< ptree::const_assoc_iterator,
ptree::const_assoc_iterator > bounds = g.equal_range( "path" );
for( ptree::const_assoc_iterator it = bounds.first; it != bounds.second; ++it )
{
std::cout << it->first << " : ";
const ptree& d = it->second.get_child( "<xmlattr>.d" );
std::cout << d.get_value<std::string>() << "\n";
}
}
}
}