使用Linq To XML，获取所有叶子路径的方法？

Ok, I've got the following XML tree

好的，我有以下XML树

<root>
    <A>
        <A1>
            <A1A>1000</A1A>
            <A1B>2000</A1B>
            <A1C>3000</A1C>
        </A1>
        <A2>
            <A2A>4000</A2A>
            <A2B>5000</A2B>
        </A2>
    </A>
    <B>
        <B1>
            <B1A>6000</B1A>
        </B1>
    </B>
</root>

From a method receiving an XDocument I want to produce a dictionary where the key is the path (really an XPath) and the value comes from the value in the corresponding leaf.

从接收XDocument的方法我想生成一个字典，其中键是路径（实际上是XPath），值来自相应叶子中的值。

root/A/A1/A1A    1000
root/A/A1/A1B    2000
root/A/A1/A1C    3000
root/A/A2/A2A    4000
root/A/A2/A2B    5000
root/B/B1/B1A    6000

Seems simple to do in Linq to XML but I can't wrap my head around it.

在Linq to XML中似乎很简单，但我无法理解它。

2 个解决方案

#1

You can find the leaves by looking for elements that have no descendants:

您可以通过查找没有后代的元素来查找叶子：

var doc = XDocument.Load(fileName);
var leaves = 
    from e in doc.Descendants()
    where !e.Elements().Any()
    select e;

I don't know if there is a built-in way to get the path of an element, but you can easily create an extension method to build it:

我不知道是否有内置的方法来获取元素的路径，但您可以轻松地创建一个扩展方法来构建它：

static class Extensions
{
    public static string Path(this XElement element)
    {
        XElement tmp = element;
        string path = string.Empty;
        while (tmp != null)
        {
            path = "/" + tmp.Name + path;
            tmp = tmp.Parent;
        }
        return path;
    }
}

You can then build the dictionary like this:

然后，您可以像这样构建字典：

var dict = leaves.ToDictionary(e => e.Path(), e => e.Value);

#2

After parsing the XML to an XDocument, which I assume you've already been able to do, use the methods below. Note that the GetPath() implementation is fairly naiive. See this answer for a better implementation.

在将XML解析为XDocument之后，我假设您已经能够执行此操作，请使用以下方法。请注意，GetPath（）实现非常简单。请参阅此答案以获得更好的实施。

public Dictionary<string, string> GetLeaves(XDocument doc)
{
    var dict = doc
        .Descendants()
        .Where(e => !e.HasElements)
        .ToDictionary(e => GetPath(e), e.Value);

    return dict;
}

private string GetPath(XElement element)
{
    var nodes = new List<string>();
    var node = element;
    while (node != null)
    {
        nodes.Add(node.Name.ToString());
        node = node.Parent;
    }

    return string.Join("/", Enumerable.Reverse(nodes));
}

#1