一个xml文件操作的问题

Question

<?xml version="1.0" encoding="gb2312"?>  <QuickAsh> 
      <Time Point = "12:15">
          <PD301>0.0</PD301>
          <CleanCoal>0.0</CleanCoal>
          <MiddleCoal>0.0</MiddleCoal>
          <Gangue>0.0</Gangue>
      </Time >
      <Time Point = "13:15">
          <PD301>35.04</PD301>
          <CleanCoal>9.05</CleanCoal>
          <MiddleCoal>26.50</MiddleCoal>
          <Gangue>73.62</Gangue>
      </Time >  </QuickAsh>现在我想得到12.15 这个属性值，求解

Answer 1

          string xml = @"<?xml version=""1.0"" encoding=""gb2312""?>  <QuickAsh> 
      <Time Point = ""12:15"">
          <PD301>0.0</PD301>
          <CleanCoal>0.0</CleanCoal>
          <MiddleCoal>0.0</MiddleCoal>
          <Gangue>0.0</Gangue>
      </Time >
      <Time Point = ""13:15"">
          <PD301>35.04</PD301>
          <CleanCoal>9.05</CleanCoal>
          <MiddleCoal>26.50</MiddleCoal>
          <Gangue>73.62</Gangue>
      </Time >  </QuickAsh>";            XDocument doc=XDocument.Parse(xml);
            var Query = from node in doc.Element("QuickAsh").Elements("Time")
                        where node.Attribute("Point").Value == "12:15"
                        select node;
            foreach (var v in Query)
            {
                Console.WriteLine("PD301={0}  CleanCoal={1}   MiddleCoal={2}  Gangue={3}",
                    v.Element("PD301").Value, v.Element("CleanCoal").Value, v.Element("MiddleCoal").Value,
                    v.Element("Gangue").Value);
            }            /*
            PD301=0.0  CleanCoal=0.0   MiddleCoal=0.0  Gangue=0.0             */

Answer 2

===============================================是不是我没表达清自己的意思：我只想取出来  time 结点下的Point属性的值：12:15或者 13:15

Answer 3

            string xml = @"<?xml version=""1.0"" encoding=""gb2312""?>  <QuickAsh> 
      <Time Point = ""12:15"">
          <PD301>0.0</PD301>
          <CleanCoal>0.0</CleanCoal>
          <MiddleCoal>0.0</MiddleCoal>
          <Gangue>0.0</Gangue>
      </Time >
      <Time Point = ""13:15"">
          <PD301>35.04</PD301>
          <CleanCoal>9.05</CleanCoal>
          <MiddleCoal>26.50</MiddleCoal>
          <Gangue>73.62</Gangue>
      </Time >  </QuickAsh>";            Console.WriteLine("LINQ TO XML");
            XDocument doc = XDocument.Parse(xml);
            var Query = from node in doc.Element("QuickAsh").Elements("Time")
                        select new { Point = node.Attribute("Point").Value };
            foreach (var v in Query)
            {
                Console.WriteLine(v.Point);
            }            //或者
            Console.WriteLine("\r\n");
            System.Xml.XmlDocument D = new System.Xml.XmlDocument();
            D.LoadXml(xml);
            foreach (System.Xml.XmlNode n in D.SelectNodes("QuickAsh/Time"))
            {
                Console.WriteLine(n.Attributes["Point"].Value);
            }
            /*
            LINQ TO XML
            12:15
            13:15
            12:15
            13:15
            */

Answer 4

You can use linq to xml to operate this xml file.

Answer 5

 ((XmlElement)doc.DocumentElement[节点名]).GetAttribute(节点属性)