7

我有以下 XML 文件，我想使用以下结构存储它。

数据结构：

    struct transitions
    {
     string oldstate;
     string event;
     string newstate;
    };

    struct XML_Diagram
    {
     string diag_name;
     string diag_defaultstate;
     list<string> diag_states;
     list<string> diag_events;
     list<transitions> diag_transitions;
    };

xml文件：

    <diagram>
      <diagname>DiagaX</diagname>
      <states>
         <state>A</state>
         .............       
      </states>
      <events>
          <event>ev1</event>
          .................
      </events>
      <defaultstate>A</defaultstate>
      <transitions>
          <transition>
              <oldstate>A</oldstate>
              <event>ev1</event>
              <newstate>B</newstate>
          </transition>
          <transition>
              <oldstate>B</oldstate>
              <event>ev2</event>
              <newstate>C</newstate>
          </transition>
          .........................
      </transitions>
    </diagram>

我很清楚如何访问 diagram.states 。我可以用以下代码做到这一点：

    using boost::property_tree::ptree;
    ptree pt;

    // Get diagram states
    BOOST_FOREACH(ptree::value_type &v, pt.get_child("diagram.states"))
    {
       diag_states.push_back(v.second.data());
    }

我不清楚的是如何从 diagram.transitions.transition 级别访问数据？

我的问题是我在文档中找不到任何关于如何解析具有多个级别的复杂 xml 文件的示例。

4

2 回答 2

6

这个有用的实用程序函数遍历并漂亮地打印整个属性树：

using boost::property_tree::ptree;

std::string q(const std::string& s) { return "\"" + s + "\""; }

void print_tree(const ptree& pt, int level)
{
    const std::string sep(2 * level, ' ');
    BOOST_FOREACH(const ptree::value_type &v, pt) {
        std::cout
            << sep
            << q(v.first) << " : " << q(v.second.data()) << "\n";
        print_tree(v.second, level + 1);
    }
}

void print_tree(const ptree& pt) { print_tree(pt, 0); }

这些v.second值是可以使用常用get方法访问的树本身。例如，可以像这样访问和打印转换：

using std::string;

void print_transitions(const ptree& pt)
{
    BOOST_FOREACH(
        const ptree::value_type &v,
        pt.get_child("diagram.transitions"))
    {
        const ptree& child = v.second;
        std::cout
            << "Event "
            << child.get<string>("event")
            << " in state "
            << child.get<string>("oldstate")
            << " leads to state "
            << child.get<string>("newstate")
            << "\n";
    }
}
于 2012-01-01T18:58:54.453 回答
4

这是另一个如何ptree使用属性打印的示例：

namespace pt = boost::property_tree;

// worker
typedef std::pair<const pt::ptree&, unsigned> tree_printer;

// empty ptree helper
const pt::ptree& empty_ptree()
{
    static pt::ptree t;
    return t;
}


std::ostream& operator <<(std::ostream& os, const tree_printer& p)
{
    const pt::ptree& tree = p.first;

    if(tree.empty()) return os;

    const std::string indent(p.second, ' ');

    BOOST_FOREACH(const pt::ptree::value_type& v, tree)
    {
        const std::string& nodeName = v.first;

        if(nodeName == "<xmlattr>") continue;

        os << indent << nodeName;
        const pt::ptree& attributes =
            v.second.get_child("<xmlattr>", empty_ptree());

        if(!attributes.empty())
        {
            os << " [ ";
            BOOST_FOREACH(const pt::ptree::value_type& attr, attributes)
            {
                const std::string& attrName = attr.first;
                const std::string& attrVal = attr.second.data();

                os << attrName << " = '" << attrVal << "'; ";
            }
            os << "]";
        }
        os << "\n";

        const pt::ptree& childNode = v.second;
        os << tree_printer(childNode, p.second + 1);
    }

    return os;
}


std::ostream& operator <<(std::ostream& os, const pt::ptree& tree)
{
    os << tree_printer(tree, 0);

    return os;
}

希望这可以帮助。

于 2012-02-06T07:30:45.660 回答