使用LINQ to XML,方法获取路径,所有的树叶? [英] Using Linq To XML, method to get path to all leaves?
问题描述
好吧,我有以下XML树
<预类=郎咸平的XML prettyprint-覆盖>
<根>
< A>
< A1,GT;
<&A1A GT; 1000℃; / A1A>
<&A1B GT; 2000< / A1B>
<&A1C GT; 3000 LT; / A1C>
< / A1>
< A2>
<&A2A GT; 4000< / A2A>
<&A2B GT; 5000< / A2B>
< / A2>
< / A>
< B>
< B1>
<&B1A GT; 6000< / B1A>
< / B1>
< / B>
< /根>
这是一个方法接受一个XDocument我想制作一本字典,其中的关键是路径(一个真正的XPath的)和值来自于相应的叶值。
根/ A / A1 / A1A 1000
根/ A / A1 / A1B 2000
根/ A / A1 / A1C 3000
根/ A / A2 / A2A 4000
根/ A / A2 / A2B 5000
根/ b / B1 / B1A 6000
看似简单Linq中做XML,但我不能换我。头周围
您可以找到通过查找元素的叶子有没有后代:
VAR DOC = XDocument.Load(文件名);
无功叶=
从E在doc.Descendants()
,其中e.Elements()任何()
选择e!。
我不知道是否有一个内置的方式来获得一个元素的路径,但是你可以很容易地创建一个扩展方法来构建它:
静态类扩展
{
公共静态字符串路径(这的XElement元素)
{
的XElement TMP =元素;
路径字符串=的String.Empty;
,而(TMP!= NULL)
{
路径=/+ tmp.Name +路径;
TMP = tmp.Parent;
}
返回路径;
}
}
您可以再建字典是这样的:
VAR字典= leaves.ToDictionary(E => e.Path(),E => e.Value);
Ok, I've got the following XML tree
<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.
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.
You can find the leaves by looking for elements that have no descendants:
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:
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:
var dict = leaves.ToDictionary(e => e.Path(), e => e.Value);
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