Question

我有一个如下所示的xml，因为我想得到第二个SubNetwork元素值和tval值，所以这里的第一行值为NC22和tet_DS_U_6359_HolidayExpressInKV 。我按照如下所示的prtial代码进行了操作，如何获取每行的第二个SubNetwork和tval值，最后我将放入tval是关键字的dictionery。这个很大几GB大小的XML，我发布了一小部分

(XmlReader xr = XmlReader.Create(path))
{
    string sitename = "";
    xr.MoveToContent();
    XNamespace xn = xr.LookupNamespace("tn");
    while (xr.Read())
    {
        while (xr.NodeType == XmlNodeType.Element && xr.NamespaceURI == tn && xr.LocalName == "mentNode" && xr.GetAttribute("id") == "TRM")
        {
            try
            {
                XElement sniipet = (XElement)XNode.ReadFrom(xr);
                var xdoc = new XDocument(sniipet);
                var orders = (from r in xdoc.Descendants(tn + "manages")
                              select new
                              {


                              }).ToList();

            }
            catch (Exception ex)
            {

            }
        }
    }
}

和XML如下所示

<?xml version="1.0" encoding="utf-8" ?>
<tn:mentNode id="TRM">
  <tn:attributes>
    <tn:manages>SubNetwork=TRM_R,SubNetwork=NC22,tval=tet_DS_U_6359_HolidayExpressInKV,Element=1</tn:manages>
    <tn:manages>SubNetwork=TRM_R,SubNetwork=NC22,tval=tet_DS_U_6352_MediaOneHtl,Element=1</tn:manages>
  </tn:attributes>
</tn:mentNode>

修改

虽然XML很大<tn:mentNode id="TRM">节点并不是我在dictionery中可以容纳的那么多

Answer 1

您只能使用XDocument而不是XmlReader：

            XDocument xr = XDocument.Load(path);

            List<string> orders = xr.Descendants().Where(x => (x.Name.LocalName == "mentNode") && ((string)x.Attribute("id") == "TRM"))
                .Select(x => x.Descendants().Where(y => y.Name.LocalName == "manages").Select(y => (string)y))
                .SelectMany(x => x).ToList();

如何从XML获取内部值

1 个答案: