Reputation: 373
I have an XML document I'm trying to deseralize that has a attribute that is ref which in C# can not be used to declare a variable hence the below doesn't work
[XmlAttribute()]
public string ref;
Anyway to get this to deseralize properly? I know it is case sensitive so Ref wouldn't work.
Upvotes: 0
Views: 194
Reputation: 136
You can change the attribute name in the xml file, by using the AttributeName, such as in the following example:
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;
namespace soans
{
public class Test
{
//problematic attribute (ref is reserved)
[XmlAttribute(AttributeName="ref")]
public string RefAttr {get;set;}
//other attributes as well
[XmlAttribute()]
public string Field { get; set; }
}
class Program
{
static void Main(string[] args)
{
string filename = ""; //use your path here
Test original = new Test()
{
RefAttr = "ref",
Field = "test"
};
//serialiser
XmlSerializer ser = new XmlSerializer(typeof(Test));
//save to file
TextWriter writer = new StreamWriter(filename);
ser.Serialize(writer, original);
writer.Close();
//read from file
TextReader reader = new StreamReader(filename);
var fromfile = ser.Deserialize(reader) as Test;
if(fromfile!=null)
{
Console.WriteLine(fromfile.RefAttr);
}
reader.Close();
Console.ReadKey();
}
}
}
Upvotes: 0
Reputation: 3311
You can provide a name in the attribute:
[XmlAttribute("ref")]
public string anynameyouwant;
Upvotes: 2