获取提升property_tree父节点 [英] Getting boost property_tree parent node

问题描述

我在程序中使用了boost property_tree。我已经将树设置为使用自定义路径类型。我正在寻找的是获取特定节点的父节点ID。

I am using boost property_tree in my program. I've set up the tree for using custom path type. What I'm looking for is getting a specific node's parent node id.

这里是一个示例：

MetaStorageTree tree;

typedef boost::property_tree::basic_ptree<Framework::CommonClientServer::InterfacePathChain_t, MetaStorageTreeNode*>
    MetaStorageTreeNode_t;
class MetaStorageTree : public MetaStorageTreeNode_t;

MetaStorageTreeNode* node = new MetaStorageTreeNode(1);
MetaStorageTreeNode* node1 = new MetaStorageTreeNode(2);
tree.put(InterfacePathChain_t{0}, node);
tree.put(InterfacePathChain_t{0, 0}, node1);
tree.put(InterfacePathChain_t{0, 1}, node1);
tree.put(InterfacePathChain_t{0, 0, 0}, node);
tree.put(InterfacePathChain_t{0, 1, 0}, node1);
tree.put(InterfacePathChain_t{0, 1, 1}, node);

//InterfacePathChain_t is basically a vector<int>

结果按预期进行：

{0}: 1
    {0}: 2
        {0}: 1
    {1}: 2
        {0}: 2
        {1}: 1

我需要的是获取完整ID的方法没有永久存储的节点。我在想的是一种简单地获取其父节点ID并将其推到路径的最前面，以此类推的方法。但是我似乎无法在property_tree中找到一种方法来执行此操作。这可能吗？如果不是，是否有其他方法可以计算这种情况下的完整路径？

What I need is a way of getting full id of a node without permanently storing it. What I am thinking of is a way of simply getting its parent node id and pushing it to the front of the path and so on to the top level. But I didn't seem to be able to find a way to do this in property_tree. Is this possible? If no, are there any other ways of calculating full path for this case?

例如，路径为{0，1，0}的节点：

For example as for node with path {0, 1, 0}:

1. id == 0 =>路径= {0}


2. parent！= NULL => parent.id = = 1 =>路径= {1，0}


3. 父母！= NULL => parent.id == 0 =>路径= {0，1，0}


4. parent == NULL => end

1. id == 0 => path = {0}
2. parent != NULL => parent.id == 1 => path = {1, 0}
3. parent != NULL => parent.id == 0 => path = {0, 1, 0}
4. parent == NULL => end

推荐答案

。

Boost Ptree节点是独立的，不知道任何包含的数据结构（这是单链列表的树等效项）。

Boost Ptree nodes are self-contained and do not know about any containing datastructure (it's the the "tree" equivalent of a singly-linked list).

作为最佳近似，您可以在父母内部查找孩子与类似 C ++：boost ptree相对键。

As a best approximation you could look up the child inside the parent, e.g. with something like in C++: boost ptree relative key.

这假定您始终可以搜索 root。

This presumes you always have "a root" available to search in.

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