Reputation: 375
Started practicing with XML and C# and I have an error message of "There is an error in XML document (3,2)". After looking at the file, I can't see anything wrong with it (Mind you, I probably missed something since I'm a noob). I'm using a Console Application for C# right now. I'm trying to return a list of Adventurers and just a side note, the GEAR element is optional. Here is what I have so far:
XML File - Test1
<?xml version="1.0" encoding="utf-8"?>
<Catalog>
<Adventurer>
<ID>001</ID>
<Name>John Smith</Name>
<Address>123 Fake Street</Address>
<Phone>123-456-7890</Phone>
<Gear>
<Attack>
<Item>
<IName>Sword</IName>
<IPrice>15.00</IPrice>
</Item>
<Item>
<IName>Wand</IName>
<IPrice>20.00</IPrice>
</Item>
</Attack>
<Defense>
<Item>
<IName>Shield</IName>
<IPrice>5.00</IPrice>
</Item>
</Defense>
</Gear>
</Adventurer>
<Adventurer>
<ID>002</ID>
<Name>Guy noone likes</Name>
<Address>Some Big House</Address>
<Phone>666-666-6666</Phone>
<Gear></Gear>
</Adventurer>
</Catalog>
C# Classes
public class Catalog
{
List<Adventurer> Adventurers { get; set; }
}
public class Adventurer
{
public int ID { get; set; }
public string Name { get; set; }
public string Address { get; set; }
public string Phone { get; set; }
public Gear Gear { get; set; }
}
public class Gear
{
public List<Item> Attack { get; set; }
public List<Item> Defense { get; set; }
}
public class Item
{
public string IName { get; set; }
public decimal IPrice { get; set; }
}
Serialize Function - Where the Problem Occurs at Line 5
Catalog obj = null;
string path = @"C:\Users\Blah\Desktop\test1.xml";
XmlSerializer serializer = new XmlSerializer(typeof(Catalog));
StreamReader reader = new StreamReader(path);
obj = (Catalog)serializer.Deserialize(reader);
reader.Close();
Console.ReadLine();
Upvotes: 5
Views: 4607
Reputation: 11
I was able to get your xml to deserialize with a couple minor changes (namely the public qualifier on Adventurer).
using System;
using System.Collections.Generic;
using System.Text;
using System.IO;
using System.Xml.Serialization;
namespace TempSandbox
{
[XmlRoot]
public class Catalog
{
[XmlElement("Adventurer")]
public List<Adventurer> Adventurers;
private readonly static Type[] myTypes = new Type[] { typeof(Adventurer), typeof(Gear), typeof(Item) };
private readonly static XmlSerializer mySerializer = new XmlSerializer(typeof(Catalog), myTypes);
public static Catalog Deserialize(string xml)
{
return (Catalog)Utils.Deserialize(mySerializer, xml, Encoding.UTF8);
}
}
[XmlRoot]
public class Adventurer
{
public int ID;
public string Name;
public string Address;
public string Phone;
[XmlElement(IsNullable = true)]
public Gear Gear;
}
[XmlRoot]
public class Gear
{
public List<Item> Attack;
public List<Item> Defense;
}
[XmlRoot]
public class Item
{
public string IName;
public decimal IPrice;
}
}
I'm using [XmlElement("Adventurer")] because the xml element names don't exactly match the class property names.
NOTE: I'm using a generic deserialization utility i already had on hand .
Upvotes: 1
Reputation: 26078
Your XML doesn't quite line up with your objects... namely these two...
public string City { get; set; }
and
<Address>123 Fake Street</Address>
Change City to Address or vice versa and it should fix the problem.
Edit: Got this to work in a test project, combination of all our answers...
Add <Adventurers>
tag after <Catalog>
(and </Adventurers>
before </Catalog>
) and change
List<Adventurer> Adventurers { get; set; }
to
public List<Adventurer> Adventurers { get; set; }
and it works properly for me.
Upvotes: 1
Reputation: 654
First, "Adventurers" property is not public, it's inaccessible, I think that the best way to find the error is to serialize your object and then compare the result with your xml file.
Upvotes: 2
Reputation: 23208
The issue is the list of Adventurers in Catalog:
<?xml version="1.0" encoding="utf-8"?>
<Catalog>
<Adventurers> <!-- you're missing this -->
<Adventurer>
</Adventurer>
...
<Adventurer>
</Adventurer>
</Adventurers> <!-- and missing this -->
</Catalog>
You don't have the wrapping element for the Adventurers
collection.
EDIT: By the way, I find the easiest way to build the XML structure and make sure it's compatible is to create the object(s) in C#, then run through the built-in XmlSerializer
and use its XML output as a basis for any XML I create rather than forming it by hand.
Upvotes: 3