XMLSerialization
本文关键字:XMLSerialization | 更新日期: 2023-09-27 17:49:56
如何序列化/反序列化以下内容
<<p> 反应/strong><xml>
<record type="user">
<field name="userId" value="Id1"/>
<field name="userName" value="Name1"/>
<field name="userStreet" value="Street1"/>
<field name="userCountry" value="Country1"/>
</record>
<record type="user">
<field name="userId" value="Id2"/>
<field name="userName" value="Name2"/>
<field name="userStreet" value="Street2"/>
<field name="userCountry" value="Country2"/>
</record>
</xml>
User.cs
class User
{
public int UserId{get;set;}
public string UserName{get;set;}
public string UserStreet{get;set;}
public string UserCountry{get;set;}
}
我不知道从哪里开始,因为我还没有遇到过这样的响应结构。
您的XML是不同的,因为对象属性是元素"field"的内部属性,而不是像
这样的元素<xml>
<User>
<UserName>
Name1
</UserName>
</User>
</xml>
所以我要做的是写一个函数来为我做解析,比如
public static User get_user(IEnumerable<XElement> fields)
{
User u = new User();
foreach (var item in fields)
{
String attribute = item.Attribute("name").Value;
String val = item.Attribute("value").Value;
if (attribute == "userName")
{
u.UserName = val;
}
else if (attribute == "userStreet")
{
u.UserStreet = val;
}
//and so on...
}
return u;
}
然后像
一样使用//using System.Xml.Linq;
XDocument document =XDocument.Parse(string_response); //I'm assuming response is in string
List<User> users = (from xnode in document.Element("xml").Elements("record")
select get_user(xnode.Elements("field"))
).ToList();
在Linq表达式中,我使用get_user()代替通常的新User{},它接受"字段"列表并返回User
试试这样
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
using System.Xml;
using System.Xml.Serialization;
namespace ConsoleApplication1
{
class Program
{
const string FILENAME = @"C:'temp'test.xml";
static void Main(string[] args)
{
Root root = new Root()
{
records = new List<Record>(){
new Record(){
type = "user",
fields = new List<Field>(){
new Field() { name= "userId", value = "Id1"},
new Field() { name= "userName", value = "Name1"},
new Field() { name="userStreet", value = "Street1"},
new Field() { name="userCountry", value="Country1"}
}
},
new Record(){
type = "user",
fields = new List<Field>(){
new Field() { name = "userId", value="Id2"},
new Field() { name = "userName", value="Name2"},
new Field() { name= "userStreet", value="Street2"},
new Field() { name="userCountry", value="Country2"}
}
}
}
};
XmlSerializer serializer = new XmlSerializer(typeof(Root));
StreamWriter writer = new StreamWriter(FILENAME);
XmlSerializerNamespaces _ns = new XmlSerializerNamespaces();
_ns.Add("", "");
serializer.Serialize(writer, root, _ns);
writer.Flush();
writer.Close();
writer.Dispose();
XmlSerializer xs = new XmlSerializer(typeof(Root));
XmlTextReader reader = new XmlTextReader(FILENAME);
Root newRoot = (Root)xs.Deserialize(reader);
List<User> users = new List<User>();
foreach (Record record in newRoot.records.AsEnumerable())
{
User newUser = new User();
users.Add(newUser);
foreach (Field field in record.fields)
{
switch (field.name)
{
case "userId" :
newUser.UserId = field.value;
break;
case "userName":
newUser.UserName = field.value;
break;
case "userStreet":
newUser.UserStreet = field.value;
break;
case "userCountry":
newUser.UserCountry = field.value;
break;
}
}
}
}
}
[XmlRoot("Root")]
public class Root
{
[XmlElement("record")]
public List<Record> records { get; set; }
}
[XmlRoot("record")]
public class Record
{
[XmlAttribute("type")]
public string type { get; set; }
[XmlElement("field")]
public List<Field> fields { get; set; }
}
[XmlRoot("field")]
public class Field
{
[XmlAttribute("name")]
public string name { get; set; }
[XmlAttribute("value")]
public string value { get; set; }
}
class User
{
public string UserId { get; set; }
public string UserName { get; set; }
public string UserStreet { get; set; }
public string UserCountry { get; set; }
}
}