我一直在使用如下文件来保存数据:
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
解决方法
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.Email); } } }
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