c# - 如果我有子节点的属性名称，如何获取父节点属性名称？

Question

嗨，我想从 xml 中获取 up 节点。例如这里是一个xml结构......

<feature name="mod1">
  <user name="user1"></user>
  <user name="user2"></user>
  <user name="user3"></user>
</feature>

我在我的应用程序中有用户名，我想要的不是节点功能名称属性。

XmlReader reader = XmlReader.Create(new StringReader(xml));

            XElement doc = XElement.Load(reader);

            int counter = 0; 
            foreach (XElement user in doc.Descendants("USER"))
            {
                try
                {
                    row = tb.NewRow();
                    row["ID"] = counter++;
                    row["Name"] = user.Attribute("NAME").Value;
                    row["Host"] = user.Attribute("HOST").Value;
                    row["Used_Licenses"] = user.Attribute("USED_LICENSES").Value;
                    row["Checkout_Time"] = user.Attribute("CHECKOUT_TIME").Value;


                   row["Modul"] = user.Parent.Attribute("NAME").Value; //don't work :(

                    tb.Rows.Add(row); 
                }
                catch (Exception)
                {

                }

            }

Answer 1

你应该能够得到它

user.Parent.Attribute("NAME").Value;

这适用于：

<LM-X STAT_VERSION="3.32">
<LICENSE_PATH >
<FEATURE NAME="GlobalZoneEU" >
<USER NAME="SYSTEM" HOST="LRV171" IP="172.16.11.115" USED_LICENSES="2000" LOGIN_TIME="2013-04-17 12:42" CHECKOUT_TIME="2013-04-17 12:42" SHARE_CUSTOM="hweuser:172.16.11.115"/>
<USER NAME="pbsadmin" HOST="SERV11" IP="172.16.11.115" USED_LICENSES="720" LOGIN_TIME="2013-04-17 12:44" CHECKOUT_TIME="2013-04-17 12:44" SHARE_CUSTOM="pbsadmin:LWSERV171:1592_40960072_1356792762_826820"/>
</FEATURE>
</LICENSE_PATH>
</LM-X>

编辑

通过用户名获取功能名称：

var featureNames = xDoc.Descendants("USER")
                        .Where(x => x.Attribute("NAME").Value == <your input>)
                        .Select(x => x.Parent.Attribute("NAME").Value);

var firstFeatureName = featureNames.FirstOrDefault();

Answer 2

似乎属性区分大小写。

XmlReader reader = XmlReader.Create(\\File Path);

        XElement doc = XElement.Load(reader);

        int counter = 0;
        foreach (XElement user in doc.Descendants("USER"))
        {
            try
            {
                string node = user.Parent.Attribute("NAME").Value; //Working - Returning 'GlobalZoneEU'
            }
            catch (Exception)
            {

            }
        }

Answer 3

以下代码应该适合您。

        string xml = "<feature name=\"mod1\">";
        xml += "<user name=\"user1\"> </user>";
        xml += "<user name=\"user2\"> </user> ";
        xml += "<user name=\"user3\"></user>";
        xml += "</feature>";

        XmlDocument xdoc=new XmlDocument();
        xdoc.LoadXml(xml);

        XDocument mydoc = XDocument.Parse(xdoc.OuterXml);

        var result = mydoc.Elements("feature").Where(parent =>
                          parent.Elements("user").Any(child =>   
                          child.Attribute("name").Value == "user2"));

并且不要忘记包含使用 System.Xml；使用 System.Xml.Linq；