我需要使用LINQ to XML查询来确定值吗？

Question

<?xml version="1.0" encoding="utf-8" ?>
<MYROOT>
  <Group name="G1">
      <Skill name="G1Skill1">
              <SubSkill>G1skill1sub1</SubSkill>
              <SubSkill>G1skill1sub2</SubSkill>
      </Skill >
    <Skill name="G1Skill2">
              <SubSkill>G1skill2sub1</SubSkill>
              <SubSkill>G1skill2sub2</SubSkill>
      </Skill >
  </Group>
  <Group name="G2">
    <Skill name="G2Skill1">
              <SubSkill>G2skill1sub1</SubSkill>
              <SubSkill>G2skill1sub2</SubSkill>
    </Skill >
    <Skill name="G2Skill2">
              <SubSkill>G2skill2sub1</SubSkill>
              <SubSkill>G2skill2sub2</SubSkill>
    </Skill >
 </Group>
</MYROOT>

例如。输入为"G1"和"G1skill1sub1"

预期输出= G1Skill1 (使用单个LINQ to XML)

我试过这段代码：

  var xmlDoc = XDocument.Load(Server.MapPath("XMLFile1.xml"));
        var skills =
        from skill in xmlDoc.Descendants("Group")
        where skill.Attribute("name").Value.Equals("G1") // && skill.Element("SubSkill").Value.Equals("Group")
        select new
            {
                Myskill = skill.Element("Skill").Value,
                mytext= skill.Attribute("name").Value,
            };

但是没有得到正确的输出。

Answer 1

我将使用：

var skills = from group in doc.Descendants("Group")
             where (string) group.Attribute("name") == "G1"
             from subskill in group.Descendants("SubSkill")
             where (string) subskill == "G1skill1sub1"
             select (string) subskill.Parent.Attribute("name");

这是根据您的输入和预期输出进行的：

• 查找所有匹配组
• 查找这些组内的所有匹配子技能
• 对于每个匹配子技能，查找父(技能)的名称

Answer 2

我试着这样做，效果很好。

         string input1 = "G1";
        string input2 = "G1skill1sub1";

        string output =
            (from skill in xdoc.Descendants("Skill")
             let subskill = skill.Element("SubSkill")
             let g = skill.Parent
             where g.Name == "Group" && g.Attribute("name").Value == input1
             && subskill != null && subskill.Value == input2
             select skill.Attribute("name").Value).FirstOrDefault();

        Console.WriteLine(output);