How to XML deserialize Dictionary, that hidden in other element?

Question

ACTUAL code and xml file

Code of program

class Program
{
    static void Main(string[] args)
    {
        string path = AppDomain.CurrentDomain.BaseDirectory + "file.xml";
        //string path2 = @"F:\fd.xml";
        FileStream fs = new FileStream(path, FileMode.Open);
XmlReader reader = XmlReader.Create(fs);
        SaveGame sav = new XmlSerializer(typeof(SaveGame)).Deserialize(reader) as SaveGame;
        Console.WriteLine(sav.player.friendshipData[0].key);
        Console.WriteLine(sav.player.friendshipData[0].value.Points);
        fs.Close();
        Console.ReadKey();
    }
}

public class SaveGame
{
    public Player player { get; set; }
}

public class Player
{
    public item[] friendshipData { get; set; }
}

public class item
{
    public string key { get; set; }
    public Friendship value { get; set; }
}

public class Friendship
{
    public int Points {get;set;}
}
}

XML File to work with:

<SaveGame>
    <player>
        <friendshipData>
            <item>
                <key>
                    <string>Name1</string>
                </key>
                <value>
                    <Friendship>
                        <Points>324</Points>
                    </Friendship>
                </value>
            </item>
            <item>
                <key>
                    <string>Name2</string>
                </key>
                <value>
                    <Friendship>
                        <Points>98</Points>
                    </Friendship>
                </value>
            </item>
        </friendshipData>
    </player>
</SaveGame>

I tried other posts, and that's not working, cause all readen values are null.

Please, how to deserialize this document? And with explanation, please.

If i set {get;set;} to variables, it won't read next item, if i set {get;} it read every item, but every item have null value.

Just in case, i can't edit XML File for some reasons. XML File is alright.

"cause i'm keeping get an exception, that means i can't deserialize this XML Document." Please be specific on what *excatly* you´ve tried and what exception you get. I doubt you want us to post the same appraoches you already tried out, do you? So help us to write *meaningful* answers that really help you. — MakePeaceGreatAgain, Jul 13 '18 at 06:58
I tried this post https://stackoverflow.com/questions/12554186/how-to-serialize-deserialize-to-dictionaryint-string-from-custom-xml-not-us — user9346149, Jul 13 '18 at 07:02
That doesn´t help much unless we know your data-structure. Please show your code of both, the types and how you de-serialize it. And again: post the **exception**. — MakePeaceGreatAgain, Jul 13 '18 at 07:04
A `key` within your `item` already *is* a string. It doesn´t have a nested string-element. Thus `...` does not represent your `item`-class. The same applies for `specificclass`. A `value` already *is* an instance of that class. There´s no need to have a further nesting in your xml for the type of a node. — MakePeaceGreatAgain, Jul 13 '18 at 07:31
that's the problem. string read normally, but specificClass always equals null — user9346149, Jul 13 '18 at 07:36
in my code, this all alright, i made a mistake on here. sorry for this. — user9346149, Jul 13 '18 at 07:38
XmlSerializer serializer = new XmlSerializer(typeof(BasicElement)); — user9346149, Jul 13 '18 at 07:43
In your xml `` is empty, so why do you expect it to be in your de-serialized object different than `null`? — MakePeaceGreatAgain, Jul 13 '18 at 07:48
it's an example xml document, in original document it has values, and class has values, that type and name correct. — user9346149, Jul 13 '18 at 07:50
Then post a [Minimal, Complete, and Verifiable example](https://stackoverflow.com/help/mcve). It´s hard to guess what goes whrong unless you give us your actual code and data. — MakePeaceGreatAgain, Jul 13 '18 at 07:54
Why should you do that? No, thas not the intended way of posting problems on Stack. Post all *relevant* information needed to understand and reproduce your issue *directly into the question*. — MakePeaceGreatAgain, Jul 13 '18 at 07:56
Now we finally have something we can work with without having to guess. — MakePeaceGreatAgain, Jul 13 '18 at 08:10
With that structure your `key`-node should be a complex object also, not just a string. Thus you need a class whose only member is the `string`-field. I´m not sure *why* having a nested `string` within your `key` even de-serialized to *anything* but leaves the element afterwards empty. That seems indeed odd. — MakePeaceGreatAgain, Jul 13 '18 at 08:20
but what if key have element name string, and i can't name variable string, as you know. — user9346149, Jul 13 '18 at 08:28
you need the verbatim to indicate you´re using a reserved keyword as field-name: `public string @string`. — MakePeaceGreatAgain, Jul 13 '18 at 08:32

score 0 · Accepted Answer · answered Jul 13 '18 at 08:50

Your data-structure doesn´t really fit your xml well. If you have a simple data-type such as int or string you can serialize and de-serialize directly into an xml-node. If you have some more complex data-structure like your Firnedship-node you need a nested node.

Having said this your data-structure should be similar to thie following:

public class SaveGame
{
    public Player player { get; set; }
}

public class Player
{
    public item[] friendshipData { get; set; }
}

public class item
{
    public Key key { get; set; }
    public Friendship value { get; set; }
}

// add this class with a single string-field
public class Key
{
    public string @string { get; set;  }
}

public class Friendship
{
    public int Points { get; set; }
}

As an aside consider following naming-conventions, which is giving classes and members of those classes PascaleCase-names, e.g. FriendshipData, Item, and Key. This however assumes you have some mapping from those names to your names within the xml. This can be done by using XmlElementAttribute:

public class Player
{
    [XmlElement("friendshipData ")] // see here how to change the name within the xml
    public item[] FriendshipData { get; set; }
}

score 0 · Answer 2 · answered Jul 13 '18 at 09:01

Since you only need two values from Xml I would not use serialization. You can get a dictionary with one linq instruction.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;

namespace ConsoleApplication51
{

    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";

        static void Main(string[] args)
        {
            XDocument doc = XDocument.Load(FILENAME);

            Dictionary<string, int> players = doc.Descendants("item")
                .GroupBy(x => (string)x.Descendants("string").FirstOrDefault(), y => (int)y.Descendants("Points").FirstOrDefault())
                .ToDictionary(x => x.Key, y => y.FirstOrDefault());

        }
    }


}

How to XML deserialize Dictionary, that hidden in other element?

2 Answers2