XML持久性不止一项？

Question

Hey I have the following code to Store the data from a list into a XML file, However when I add a 2nd Item to the list it just overwrites the first one in XML so there is only ever one Item in the XML file, How can I solve this

class

public class Visits
{
/*
 * This class represents a single appointment
 */

    private string Customer_Name;
    private string Customer_Address;
    private DateTime Arrival_Time;
    private string Visit_Type;
    private Double Lat1;
    private Double Long1;
    //Private methods. Note the use of DateTime to store arrival time

    public string name{
        //Description property
        set { Customer_Name = value; }
        get {return Customer_Name;}
    }

    public string address
    {//Time property
        set { Customer_Address = value; }
        get { return Customer_Address; }
    }

    public DateTime arrival
    {   //Duration property
        set { Arrival_Time = value; }
        get { return Arrival_Time; }
    }

    public string type
    {
        set { Visit_Type = value; }
        get { return Visit_Type; }
    }

    public Double Lat
    {
        //Description property
        set { Lat1 = value; }
        get { return Lat1; }
    }

    public Double Lon1
    {
        //Description property
        set { Long1 = value; }
        get { return Long1; }
    } 
    public override string ToString()
    {   //Return a String representing the object
        return Visit_Type + "     " + Customer_Name + " " + Customer_Address + " " + Arrival_Time.ToString() + " " + "Latitude  " + Lat1 + " " + "Longitude  " + Long1;
    }
}

}

then the list

class List
{
/*
 * This object represents the List. It has a 1:M relationship with the Visit class
 */

    private List<Visits> visits = new List<Visits>();
    //List object use to implement the relationshio with Visits

    public void addVisits(Visits vis)
    {
        //Add a new Visit to the List
        visits.Add(vis);
    }

    public List<String> listVisits()
    {//Generate a list of String objects, each one of which represents a Visit in List.

        List<String> listVisits = new List<string>();
        //This list object will be populated with Strings representing the Visits in the lists

        foreach (Visits vis in visits)
        {
            String visAsString = vis.ToString();
            //Get a string representing the current visit object

            listVisits.Add(visAsString);
            //Add the visit object to the List
        }

        return listVisits;
        //Return the list of strings
    }

    public Visits getVisits(int index)
    {
        //Return the visit object at the <index> place in the list

        int count = 0;
        foreach (Visits vis in visits)
        {
            //Go through all the visit objects
            if (index == count)
                //If we're at the correct point in the list...
                return vis;
            //exit this method and return the current visit
            count++;
            //Keep counting
        }
        return null;
        //Return null if an index was entered that could not be found
    }
}

}

then the code to add

            thePickup.name = txtCustName.Text;
            thePickup.address = txtCustAddress.Text;
            thePickup.arrival = DateTime.Parse(txtArrival.Text);
            thePickup.Dname = txtDeliveryName.Text;
            thePickup.Daddress = txtDaddress.Text;
            thePickup.Lat = Double.Parse(txtLat.Text);
            thePickup.Lon1 = Double.Parse(txtLong.Text);
            thePickup.type = "Pickup";
            //Update thePickup object to reflect any changes made by the user

            XmlSerializer SerializerObj = new XmlSerializer(typeof(Pickups));

        using (TextWriter WriteFileStream = new StreamWriter(@"Pickup.xml", true))
        {
            SerializerObj.Serialize(WriteFileStream, thePickup);
        }

When I add a new entry It just changes the format of the original entry

Answer 1

You should try to use XPath supported libraries to handle XML Assembling and creation.

As you are using C# i would recomend HtmlAgilityPack for handling the "Parsing" thing.

In order to build the XML, here is a good source for learning how to do so using XPath.

You can also use the native XMLDocument class from C# with in your case may be more useful if you are building it without any parsing logic before.

Take a look here

XPath Sample:

Here is an XPath sample that would help you to avoid overriding the actual node in the XML File.

CreateTag("//NODE1/NODE2", "TAG_NAME", tagContent);


    private void CreateTag(string xPath, string tag, string tagContent)
    {
        XmlNode node = _xml.SelectSingleNode(xPath);
        XmlElement element = _xml.CreateElement(tag);

        element.InnerText = tagContent;
        node.AppendChild(element);
    }

In case your set has multiple nodes with the same name :

CreateTag("//SINTEGRA//SEARCH//RECORDS//RECORD[last()]", kv.Key, kv.Value);

Where last() is a XPath method that is implemented by most of .dlls that returns the index to the lastnode + 1 so that your node will be inserted after the last created one

Answer 2

Try this:

using(TextWriter WriteFileStream = new StreamWriter(@"Pickups.xml", true))
{
    SerializerObj.Serialize(WriteFileStream, thePickup);
}

instead.

The boolean true parameter means that the StreamWriter will append the next written block to the existing file, instead of overwriting it

Otherwise, your code overwrites Pickups.xml file each time. And don't forget to close WriteFileStream object.

I tried to reproduce your case:

    public class Pickups
    {
        public string name { get; set; }
        public string address { get; set; }
        public string Dname { get; set; }
        public string Daddress { get; set; }
        public string type { get; set; }
        public DateTime arrival { get; set; }
        public DateTime Lat { get; set; }
        public DateTime Lon1 { get; set; }
    }


class Program
{


    static void Main()
    {
        Pickups thePickup = new Pickups();
        thePickup.name = "nameProp";
        thePickup.address = "addressProp";
        thePickup.arrival = DateTime.Now;
        thePickup.Dname = "txtDeliveryName";
        thePickup.Daddress = "txtDaddress";
        thePickup.Lat = DateTime.Now;
        thePickup.Lon1 = DateTime.Now;
        thePickup.type = "Pickup";
        //Update thePickup object to reflect any changes made by the user

        XmlSerializer SerializerObj = new XmlSerializer(typeof(Pickups));

        using (TextWriter WriteFileStream = new StreamWriter(@"Pickups.xml", true))
        {
            SerializerObj.Serialize(WriteFileStream, thePickup);
        }

        Pickups thePickup1 = new Pickups();
        thePickup1.name = "nameProp2";
        thePickup1.address = "addressProp2";
        thePickup1.arrival = DateTime.Now;
        thePickup1.Dname = "txtDeliveryName2";
        thePickup1.Daddress = "txtDaddress2";
        thePickup1.Lat = DateTime.Now;
        thePickup1.Lon1 = DateTime.Now;
        thePickup1.type = "Pickup2";

        using (TextWriter WriteFileStream = new StreamWriter(@"Pickups.xml", true))
        {
            SerializerObj.Serialize(WriteFileStream, thePickup1);
        }
    }

}

And in Pickups.xml file I get expected result(2 entities):

<?xml version="1.0" encoding="utf-8"?>
<Pickups xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <name>nameProp</name>
  <address>addressProp</address>
  <Dname>txtDeliveryName</Dname>
  <Daddress>txtDaddress</Daddress>
  <type>Pickup</type>
  <arrival>2012-12-05T15:30:37.809487+01:00</arrival>
  <Lat>2012-12-05T15:30:37.810487+01:00</Lat>
  <Lon1>2012-12-05T15:30:37.810487+01:00</Lon1>
</Pickups><?xml version="1.0" encoding="utf-8"?>
<Pickups xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <name>nameProp2</name>
  <address>addressProp2</address>
  <Dname>txtDeliveryName2</Dname>
  <Daddress>txtDaddress2</Daddress>
  <type>Pickup2</type>
  <arrival>2012-12-05T15:30:37.989487+01:00</arrival>
  <Lat>2012-12-05T15:30:37.989487+01:00</Lat>
  <Lon1>2012-12-05T15:30:37.989487+01:00</Lon1>
</Pickups>

Are you sure you fixed all parts of your program? Maybe you write to the same file from different places in your code?

Answer 3

Do not serialize a single element but a list:

 List<Pickups> list = new List<Pickups>();

 foreach ( var pickup in ... )
    list.Add( pickup );

 ...
 XmlSerializer SerializerObj = new XmlSerializer(typeof(List<Pickups>));

 TextWriter WriteFileStream = new StreamWriter(@"Pickups.xml");
 SerializerObj.Serialize( WriteFileStream, list );