这是我的示例XML文件:
<Main>
<Person>
<Name>Božena</Name>
<Surname>Němcová</Surname>
<Gender>Female</Gender>
<OrderNum>18</OrderNum>
<BirthDate>04.02.1820</BirthDate>
</Person>
<Person>
<Name>Jan</Name>
<Surname>Žižka</Surname>
<Gender>Male</Gender>
<OrderNum>7</OrderNum>
<BirthDate>19.09.1360</BirthDate>
</Person>
<Person>
<Name>Che</Name>
<Surname>Guevara</Surname>
<Gender>Male</Gender>
<OrderNum>27</OrderNum>
<BirthDate>14.06.1928</BirthDate>
</Person>
<Person>
<Name>Antonie</Name>
<Surname>de Saint-Exupéry</Surname>
<Gender>Male</Gender>
<OrderNum>15</OrderNum>
<BirthDate>29.06.1900</BirthDate>
</Person>
</Main>
这是我想用来获取 Name 元素的所有值的列表的代码:
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("PersonWrite.xml");
XmlNodeList data = xmlDoc.SelectNodes("Main/Person/Name");
问题是我只能从第一个 Person 元素中获取一个值。
答案 0 :(得分:2)
我喜欢使用反序列化,使用起来更容易。
using System;
using System.Xml.Serialization;
using System.IO;
public class Main
{
[XmlElement("Person")]
public Person[] People { get; set; }
}
public class Person
{
public string Name { get; set; }
public string Surname { get; set; }
public string Gender { get; set; }
public int OrderNum { get; set; }
public string BirthDate { get; set; }
}
public class Program
{
public static void Main()
{
var xmlString = @"<Main>
<Person>
<Name>Božena</Name>
<Surname>Němcová</Surname>
<Gender>Female</Gender>
<OrderNum>18</OrderNum>
<BirthDate>04.02.1820</BirthDate>
</Person>
<Person>
<Name>Jan</Name>
<Surname>Žižka</Surname>
<Gender>Male</Gender>
<OrderNum>7</OrderNum>
<BirthDate>19.09.1360</BirthDate>
</Person>
<Person>
<Name>Che</Name>
<Surname>Guevara</Surname>
<Gender>Male</Gender>
<OrderNum>27</OrderNum>
<BirthDate>14.06.1928</BirthDate>
</Person>
<Person>
<Name>Antonie</Name>
<Surname>de Saint-Exupéry</Surname>
<Gender>Male</Gender>
<OrderNum>15</OrderNum>
<BirthDate>29.06.1900</BirthDate>
</Person>
</Main>";
var serializer = new XmlSerializer(typeof (Main));
Main main = null;
using (var reader = new StringReader(xmlString))
{
main = (Main)serializer.Deserialize(reader);
}
if (main == null)
{
return;
}
Console.WriteLine(main.People.Length);
}
}
输出:
4
答案 1 :(得分:0)
我将遍历主节点的子节点
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("PersonWrite.xml");
XmlNodeList root = xmlDoc.SelectNodes("Main");
foreach (XmlNode xnode in root.ChildNodes)
{
//get data from xnode
}