我一直在使用如下文件来保存数据: field1 field2 field3 field4myname myhashedpass myemail@email.com more stuff afteretc etc etc etc 每行都转换为字符串(名称,通行证,电子邮件) 我想将我的文本文件(见上文
field1 field2 field3 field4 myname myhashedpass myemail@email.com more stuff after etc etc etc etc
每行都转换为字符串(名称,通行证,电子邮件)
我想将我的文本文件(见上文)转换为XML文件,如下所示:
<person1> <name>myname</name> <pass>myhashedpass</pass> <email>etc</email> </person1> <person2> etc etc etc etc
基本上,我坚持如何进行此迁移,并以与处理文本数据相同的方式操纵XML数据
对你的问题的回答是这样的:using System;
using System.Linq;
using System.Xml.Linq;
namespace XmlSerialization
{
class Program
{
static void Main(string[] args)
{
var person1 = new Person();
person1.Name = "Joe";
person1.Password = "Cla$$ified";
person1.Email = "none@your.bussiness";
var person2 = new Person();
person2.Name = "Doe";
person2.Name = "$ecret";
person2.Email = "dont@spam.me";
var persons = new[] {person1, person2};
XElement xml = new XElement("persons",
from person in persons
select new XElement("person",
new XElement("name", person.Name),
new XElement("password", person.Password),
new XElement("email", person.Email))
);
xml.Save("persons.xml");
XElement restored_xml = XElement.Load("persons.xml");
Person[] restored_persons = (from person in restored_xml.Elements("person")
select new Person
{
Name = (string)person.Element("name"),
Password = (string)person.Element("password"),
Email = (string)person.Element("email")
})
.ToArray();
foreach (var person in restored_persons)
{
Console.WriteLine(person.ToString());
}
Console.ReadLine();
}
}
public class Person
{
public string Name { get; set; }
public string Password { get; set; }
public string Email { get; set; }
public override string ToString()
{
return string.Format("The person with name {0} has password {1} and email {2}",
this.Name, this.Password, this.Email);
}
}
}
但是,让内置的serializattion类为您进行XML的转换更好.下面的代码需要显式引用System.Runtime.Serialization.dll. using语句本身是不够的:
using System;
using System.IO;
using System.Linq;
using System.Xml.Linq;
using System.Runtime.Serialization;
namespace XmlSerialization
{
class Program
{
static void Main(string[] args)
{
var person1 = new Person();
person1.Name = "Joe";
person1.Password = "Cla$$ified";
person1.Email = "none@your.bussiness";
var person2 = new Person();
person2.Name = "Doe";
person2.Name = "$ecret";
person2.Email = "dont@spam.me";
var persons = new[] {person1, person2};
DataContractSerializer serializer=new DataContractSerializer(typeof(Person[]));
using (var stream = new FileStream("persons.xml", FileMode.Create, FileAccess.Write))
{
serializer.WriteObject(stream,persons);
}
Person[] restored_persons;
using (var another_stream=new FileStream("persons.xml",FileMode.Open,FileAccess.Read))
{
restored_persons = serializer.ReadObject(another_stream) as Person[];
}
foreach (var person in restored_persons)
{
Console.WriteLine(person.ToString());
}
Console.ReadLine();
}
}
[DataContract]
public class Person
{
[DataMember]
public string Name { get; set; }
[DataMember]
public string Password { get; set; }
[DataMember]
public string Email { get; set; }
public override string ToString()
{
return string.Format("The person with name {0} has password {1} and email {2}",
this.Name, this.Password, this.Email);
}
}
}
