Dynamically Bind XML to DataGrid in Silverlight

Question

I\'ve been trying to bind XML (via an XElement) to a DataGrid dynamically in Silverlight (specifically Silverlight 4, but any solutions in SL3 would be fine too) but have be

Answer 1

Below is another alternative that may also help. It's a bit of a hack.

It's written and tested using Silverlight 3.

The ViewModel:

namespace DatagridXml
{
    public class TestViewModel
    {
        public TestViewModel()
        {
            XmlData = @"Name121
Address1Name222
Address2Name323
Address3";
        }
        public string XmlData { get; set; }
    }
}

The value converter:

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;
using System.Windows.Controls;
using System.Windows.Data;
using System.Xml.Linq;

namespace DatagridXml
{
    public class XmlColumnConverter : IValueConverter
    {
        public object Convert(object value, Type targetType, object parameter, CultureInfo culture)
        {
            string elementToGenerate = parameter.ToString();
            DataGrid control = value as DataGrid;
            control.Columns.Clear();

            var result = new List>();
            XDocument xmlDoc = XDocument.Parse(control.DataContext.ToString());

            // Generate Columns
            var columnNames = xmlDoc.Descendants(elementToGenerate).FirstOrDefault();
            int pos = 0;
            foreach (var columnName in columnNames.Elements())
            {
                var column = new DataGridTextColumn();
                column.Header = columnName.Name;
                column.Binding = new Binding("[" + pos + "]");
                control.Columns.Add(column);
                pos++;
            }

            // Parse elements to generate column's data
            foreach (var element in xmlDoc.Descendants(elementToGenerate))
            {
                var row = new List();
                foreach (var column in element.Elements())
                {
                    row.Add(column.Value);
                }
                result.Add(row);
            }
            return result;
        }

        public object ConvertBack(object value, Type targetType, object parameter, CultureInfo culture)
        {
            throw new NotSupportedException("Cannot convert to xml from list.");
        }
    }
}

And, you use like this: