XML的读取

Question

把xml读到DataSet，然后根据参数操作DataSet就行了

Answer 1

[Test]
        public void Test()
        {
            XmlDocument doc = new XmlDocument();
            doc.LoadXml(@"<images>
  <image>
    <id>0</id>
    <name>a</name>
    <namecn>jkk</namecn>
    <path>10</path>
  </image>
  <image>
    <id>1</id>
    <name>b</name>
    <namecn>sas</namecn>
    <path>10</path>
  </image>
</images>");
            Console.Write(GetStrNameCn(doc, "b")); //结果sas        }        public string GetStrNameCn(XmlDocument doc ,string strname)
        {
            //根据传入的参数strname---name节点
            //返回对应的namecn的值            XmlNode node = doc.SelectSingleNode(String.Format("//name[.=\"{0}\"]",strname));
            if(node == null) return null;
            return node.NextSibling.InnerText;
        }

Answer 2

XmlNode node = doc.SelectSingleNode(String.Format("//name[.=\"{0}\"]",strname));
这句的意思，是用xpath，找所有name节点，且内容为strname的节点

Answer 3

 public static string GetStrNameCn(string strname)
        {
            XDocument doc = XDocument.Load("../../xml.xml");
            return doc.Descendants("image").FirstOrDefault(s => s.Element("name").Value == strname).Element("namecn").Value;
        }

Answer 4

        static string GetStrName(string strname)
        {
            string xml = @"<images>
  <image>
    <id>0</id>
    <name>a</name>
    <namecn>jkk</namecn>
    <path>10</path>
  </image>
  <image>
    <id>1</id>
    <name>b</name>
    <namecn>sas</namecn>
    <path>10</path>
  </image>
</images>";
            return XDocument.Parse(xml).Descendants("image").Where(x => x.Element("name").Value == strname).Select(x => x.Element("namecn").Value).First();
        }