How to deserialize XML file and then serialize again adding new item to this file?
I need to deserialize xml file to List or Array then enlarge this List or Array on 1, then serialize again file adding new object to this XML file.
I wrote something like that but it's don't work:
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开发者_Python百科.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);
}
}
}
}
thanks for any help:)
Your deserialization code is effectively a no-op - you're not returning anything. Change it to return the deserialized list, like this:
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
}
}
then within the click event, something like this:
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);
}
}
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