rapidxml：如何遍历节点？留下最后的兄弟姐妹 [英] rapidxml: how to iterate through nodes? Leaves out last sibling

问题描述

使用rapidxml我想循环一组节点，并且使用我发现的最好的方法（从可靠的stackoverflow，doc似乎没有迭代的例子）：

Using rapidxml I'm wanting to loop through a set of nodes, and am using what I found to be the best way to do this (from trusty stackoverflow, the doc doesn't appear to have an example of iteration):

while (curNode->next_sibling() !=NULL ) {
    string shiftLength = curNode->first_attribute("shiftLength")->value();
    cout << "Shift Length " << "\t" << shiftLength << endl;
    curNode = curNode->next_sibling();        
}

不幸的是，在我的OSX 10.6上，这是遗漏了最后一个兄弟node - 我猜是因为在循环的最后一次迭代中，next_sibling被调用两次。如果我在循环之后写，我可以到达最后一个节点：

Unfortunately, on my OSX 10.6 this is leaving out the last sibling node - I guess because in the last iteration of the loop, next_sibling is called twice. I can get at this last node if I write, after the loop:

cout << " LAST IS: " << curNode->first_attribute("shiftLength")->value();

...但这很狡猾，程序在此时退出。

...but that's dodgy, and the program quits at that point.

第一个问题：这可能是我设置的唯一缺陷（OSX 10.6）还是编码错误？

First question: Could this be a unique foible of my setup (OSX 10.6) or have I coded wrong?

第二个问题：是吗任何人都有一个他们认为使用rapidxml迭代未知数量的XML节点的正确方法的例子吗？

Second question: Does anyone have an example of what they believe is the correct way to iterate through an unknown number of XML nodes using rapidxml?

谢谢你们

Pete

推荐答案

这是在rapidxml中迭代节点的所有子节点的正确方法： / p>

This is the proper way to iterate though all child nodes of a node in rapidxml:

xml_node<> *node = ...
for (xml_node<> *child = node->first_node(); child; child = child->next_sibling())
{
    // do stuff with child
}

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