如何反序列化XML文件，然後序列化再次添加新項目到這個文件？

Question

我需要將xml文件反序列化爲List或Array，然後將此List或Array放大1，然後再次序列化文件，將新對象添加到此XML文件。

我寫了類似的東西，但它不工作：

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Web; 
using System.Web.UI; 
using System.Web.UI.WebControls; 
using System.Xml; 
using System.IO; 
using System.Xml.Serialization; 

public partial class _Default : System.Web.UI.Page 
{ 
    protected void Page_Load(object sender, EventArgs e) 
    { 

    } 
    public void deserialize() 
    { 
     string path = Server.MapPath(Request.ApplicationPath + "/test.xml"); 
     using (FileStream fs = new FileStream(path, FileMode.Open)) 
     { 
      XmlSerializer ser = new XmlSerializer(typeof(List<Person>)); 
      List<Person> persons = (List<Person>)ser.Deserialize(fs); 
      fs.Close(); 
      //List<Person> persons1 = ((List<Person>)os.Length + 1); 
     } 
    } 
    protected void Button1_Click(object sender, EventArgs e) 
    { 
     string path = Server.MapPath(Request.ApplicationPath + "/test.xml"); 
     if (File.Exists(path)) 
     { 
      deserialize(); 
     } 
     else 
     { 
      List<Person> o = new List<Person>(TextBox1.Text, TextBox2.Text, int.Parse(TextBox3.Text)); 
      using (FileStream fs = new FileStream(path, FileMode.Create)) 
      { 
       XmlSerializer ser = new XmlSerializer(typeof(List<Person>)); 
       ser.Serialize(fs, o); 
      } 
     } 
    } 
} 

感謝您的幫助:)

Answer 1

你反序列化的代碼實際上是一個無操作 - 你不返回任何東西。將其更改爲回報反序列化的列表，像這樣：

private List<Person> Deserialize(string path) 
{ 
    using (FileStream fs = new FileStream(path, FileMode.Open)) 
    { 
     XmlSerializer ser = new XmlSerializer(typeof(List<Person>)); 
     return (List<Person>) ser.Deserialize(fs);//There is an error in XML document (2, 2). this error i got here 
    } 
} 

然後單擊事件中，這樣的事情：

protected void Button1_Click(object sender, EventArgs e) 
{ 
    string path = Server.MapPath(Request.ApplicationPath + "/test.xml"); 
    List<Person> people = File.Exists(path) ? Deserialize(path) 
              : new List<Person>(); 
    people.Add(new Person(TextBox1.Text, TextBox2.Text, 
          int.Parse(TextBox3.Text)); 
    using (FileStream fs = File.OpenWrite(path)) 
    { 
     XmlSerializer ser = new XmlSerializer(typeof(List<Person>)); 
     ser.Serialize(fs, o); 
    } 
}