C# 更好的读取xml的方法

我有一个xml文件

<root>
  <child_1>
    <sub_child attr1="text" attr2="text" />
    <sub_child attr1="text" attr2="text" />
    <sub_child attr1="text" attr2="text" />
  </child_1>    
  <child_2>
    <sub_child attr1="text" attr2="text" />
    <sub_child attr1="text" attr2="text" />
    <sub_child attr1="text" attr2="text" />
  </child_2>    
  <child_3>
    <sub_child_1 attr1="text" attr2="text" />
    <sub_child_2 attr1="text" attr2="text" />
  </child_3>    
  ...

我需要为child_1和child_2列出两个SubChild对象，为sub_child_1和sub_child_2列出两个SubChild对象

现在我使用LINQtoXML，我的代码是

// List<SubChild> 
var child_1 = from s in doc.Descendants("sub_child").Where(x => x.Parent.Name == "child_1")
              select new SubChild {
                a1 = s.Attribute("attr1").Value,
                a2 = s.Attribute("attr2").Value,
              };

// List<SubChild> 
var child_2 = from s in doc.Descendants("sub_child").Where(x => x.Parent.Name == "child_2")
              select new SubChild {
                a1 = s.Attribute("attr1").Value,
                a2 = s.Attribute("attr2").Value,
              };

// SubChild
var sub_1 = (from s in doc.Descendants("sub_child_1")
              select new SubChild {
                a1 = s.Attribute("attr1").Value,
                a2 = s.Attribute("attr2").Value,
              }).First();

// SubChild
var sub_2 = (from s in doc.Descendants("sub_child_2")
              select new SubChild {
                a1 = s.Attribute("attr1").Value,
                a2 = s.Attribute("attr2").Value,
              }).First();

---

但是它看起来很难看，我想问有没有更清晰的方法来实现这一点？

您可以使用.Element和.Elements指定路径，从而消除where lambda

 // untested
 var child_1 = from s in doc.Root.Element("child_1").Elements("sub_child")
          select new SubChild {
            a1 = s.Attribute("attr1").Value,
            a2 = s.Attribute("attr2").Value,
          };

---

再进一步，您可以创建一个包含元素所有子元素的字典。这将允许您通过名称直接访问每个元素的子项

Dictionary<string, List<SubChild>> dict = doc.Root.Elements().ToDictionary(
    e => e.Name.ToString(),
    e => e.Elements("sub_child")
          .Select(s => new SubChild
                       {
                           a1 = s.Attribute("attr1").Value,
                           a2 = s.Attribute("attr2").Value,
                       }).ToList());
dict["child_1"][0] // get the first subchild of "child_1"

---

您真的必须使用linq转换xml吗

public IEnumerable<SubChild> GetSubChilds(XmlDocument xd)
{
   foreach (XmlNode subChild in xd.SelectNodes("/root/*/*"))
   {
       if (subChild.Name.StartsWith("sub_child"))
       {
           XmlAttributeCollection atts = subChild.Attributes;
           yield return new SubChild { a1 = atts["attr1"].Value, a2 = atts["attr2"].Value };
       }
   }
}

---

但这比林克好吗？请注意，您将需要其中两个参数。想象一下道路的变化。

public IEnumerable<SubChild> GetSubChilds(XmlDocument xd)
{
   foreach (XmlNode subChild in xd.SelectNodes("/root/*/*"))
   {
       if (subChild.Name.StartsWith("sub_child"))
       {
           XmlAttributeCollection atts = subChild.Attributes;
           yield return new SubChild { a1 = atts["attr1"].Value, a2 = atts["attr2"].Value };
       }
   }
}