如何迭代boost属性树?

时间:2011-01-03 17:37:19

标签: c++ xml iterator boost-propertytree

我知道正在接近提升属性树,并发现它是用于c ++编程的boost库的一个很好的特性。

嗯,我有一个疑问?如何使用迭代器或类似方法迭代属性树?

在参考文献中,只有一个浏览树的例子:

BOOST_FOREACH

但是没有更多吗?像stl一样的容器?这是一个更好的解决方案,谈论代码质量....

5 个答案:

答案 0 :(得分:25)

这是我经过多次实验后想出来的。我想在社区分享它,因为我找不到我想要的东西。每个人似乎只是发布了增强文档的答案,我发现这些文档是不够的。无论如何:

#include <boost/property_tree/ptree.hpp>
#include <boost/property_tree/json_parser.hpp>
#include <string>
#include <iostream>

using namespace std; 
using boost::property_tree::ptree; 

string indent(int level) {
  string s; 
  for (int i=0; i<level; i++) s += "  ";
  return s; 
} 

void printTree (ptree &pt, int level) {
  if (pt.empty()) {
    cerr << "\""<< pt.data()<< "\"";
  }

  else {
    if (level) cerr << endl; 

    cerr << indent(level) << "{" << endl;     

    for (ptree::iterator pos = pt.begin(); pos != pt.end();) {
      cerr << indent(level+1) << "\"" << pos->first << "\": "; 

      printTree(pos->second, level + 1); 
      ++pos; 
      if (pos != pt.end()) {
        cerr << ","; 
      }
      cerr << endl;
    } 

   cerr << indent(level) << " }";     
  }

  return; 
}

int main(int, char*[]) {

  // first, make a json file:
  string tagfile = "testing2.pt"; 
  ptree pt1;
  pt1.put("object1.type","ASCII");  
  pt1.put("object2.type","INT64");  
  pt1.put("object3.type","DOUBLE");  
  pt1.put("object1.value","one");  
  pt1.put("object2.value","2");  
  pt1.put("object3.value","3.0");  
  write_json(tagfile, pt1); 

  ptree pt;
  bool success = true; 

  try {
      read_json(tagfile, pt); 
      printTree(pt, 0); 
      cerr << endl; 
  }catch(const json_parser_error &jpe){
      //do error handling
      success = false
  }

  return success; 
}

这是输出:

rcook@rzbeast (blockbuster): a.out
{
  "object1": 
  {
    "type": "ASCII",
    "value": "one"
   },
  "object2": 
  {
    "type": "INT64",
    "value": "2"
   },
  "object3": 
  {
    "type": "DOUBLE",
    "value": "3.0"
   }
 }
rcook@rzbeast (blockbuster): cat testing2.pt 
{
    "object1":
    {
        "type": "ASCII",
        "value": "one"
    },
    "object2":
    {
        "type": "INT64",
        "value": "2"
    },
    "object3":
    {
        "type": "DOUBLE",
        "value": "3.0"
    }
}

答案 1 :(得分:16)

BOOST_FOREACH只是一种方便的迭代方式,可以通过iterator,begin()和end()

完成
Your_tree_type::const_iterator end = tree.end();
for (your_tree_type::const_iterator it = tree.begin(); it != end; ++it)
    ...

在C ++ 11中它是:

for (auto it: tree)
    ...

答案 2 :(得分:7)

我最近遇到了这个问题,发现答案不完整,因此我提出了这个简短而又甜蜜的片段:

using boost::property_tree::ptree;

void parse_tree(const ptree& pt, std::string key)
{
  std::string nkey;

  if (!key.empty())
  {
    // The full-key/value pair for this node is
    // key / pt.data()
    // So do with it what you need
    nkey = key + ".";  // More work is involved if you use a different path separator
  }

  ptree::const_iterator end = pt.end();
  for (ptree::const_iterator it = pt.begin(); it != end; ++it)
  {
    parse_tree(it->second, nkey + it->first);
  }
}

需要注意的是,除根节点外的任何节点都可以包含数据和子节点。 if (!key.empty())位将获取除根节点以外的所有数据,我们也可以开始构建节点子节点循环的路径(如果有的话)。

你可以通过调用parse_tree(root_node, "")开始解析,当然你需要在这个函数中做一些事情才能让它值得做。

如果您正在进行一些不需要FULL路径的解析,只需删除nkey变量及其操作,然后将it->first传递给递归函数。

答案 3 :(得分:0)

基于BFS的打印ptree遍历,如果我们要进行一些算法操作,则可以使用

int print_ptree_bfs(ptree &tree) {
try {
    std::queue<ptree*> treeQ;
    std::queue<string> strQ;

    ptree* temp;

    if (tree.empty())
        cout << "\"" << tree.data() << "\"";

    treeQ.push(&tree);
    //cout << tree.data();
    strQ.push(tree.data());

    while (!treeQ.empty()) {
        temp = treeQ.front();
        treeQ.pop();

        if (temp == NULL) {
            cout << "Some thing is wrong" << std::endl;
            break;
        }
        cout << "----- " << strQ.front() << "----- " << std::endl;
        strQ.pop();

        for (auto itr = temp->begin(); itr != temp->end(); itr++) {
            if (!itr->second.empty()) {
                //cout << itr->first << std::endl;
                treeQ.push(&itr->second);
                strQ.push(itr->first);
            } else {
                cout<<itr->first << " " << itr->second.data() << std::endl;
            }
        }

        cout << std::endl;

     }
   } catch (std::exception const& ex) {
    cout << ex.what() << std::endl;
   }
   return EXIT_SUCCESS;
  }

答案 4 :(得分:0)

答案How to iterate a boost property tree?的补充:

在基于for (auto node : tree)的C ++ 11样式范围中,每个node是一个std::pair<key_type, property_tree>

在手动编写的迭代中

Your_tree_type::const_iterator end = tree.end();
for (your_tree_type::const_iterator it = tree.begin(); it != end; ++it)
...

迭代器it是指向该对的指针。使用上的差别很小。例如,要访问密钥,可以写it->first但写node.first

发布为新答案,因为我对原始答案的建议编辑被拒绝并建议发布新答案。