如何将Dapper查询结果绑定到WPF DataGrid

Question

Here's my code. It produces a bound grid with the correct number of rows, however the cells are empty .

XAML

<DataGrid
  Name="grid" 
  ItemsSource="{Binding}"
  AutoGenerateColumns="True" />

Code behind

grid.DataContext = cn.Query("select * from SomeTable");

Answer 1

From the docs , your syntax there -- cn.Query("sql") -- returns a list of dynamic-typed objects ( IEnumerable<dynamic> ). That won't work for the DataGrid auto-columns, which looks for concrete members to generate its columns. I'd suggest creating a simple entity class for SomeTable to map the properties and then using cn.Query<SomeTableEntity>("select * from SomeTable"); .

Answer 2

So I guess the answer is: it's not possible. Here's a hacky workaround.

        ...
        var items = cn.Query("select * from SomeTable");
        grid.DataContext = ConvertToDataTable(items);
    }

    public DataTable ConvertToDataTable(IEnumerable<dynamic> items) {
        var t = new DataTable();
        var first = (IDictionary<string, object>)items.First();
        foreach (var k in first.Keys)
        {
            var c = t.Columns.Add(k);
            var val = first[k];
            if (val != null) c.DataType = val.GetType();
        }

        foreach (var item in items)
        {
            var r = t.NewRow();
            var i = (IDictionary<string, object>)item;
            foreach (var k in i.Keys)
            {
                var val = i[k];
                if (val == null) val = DBNull.Value;
                r[k] = val;
            }
            t.Rows.Add(r);
        }
        return t;
    }

Answer 3

If you use the non-generic version of Query, it returns a dynamic representation of the data. The dynamic API is not suitable for most UI data-bindings. It would be preferable to use the generic Query<T> API to load the data into types that have defined properties.

At a complete push , it would also theoretically be possible to implement ITypedList on the data and expose the properties accordingly. But that is quite a lot of work for not so much gain.

Answer 4

It is not possible with pure Linq ! It is possible reusing OLD dataset and convert back to linq using System.Data.DataSetExtensions. (Not Elegant but it works)

// Steal concrete connection from Linq
ModelDEmoWPFContainer l_ctx = new ModelDEmoWPFContainer();
var l_connection = (System.Data.EntityClient.EntityConnection)l_ctx.Connection)
                                   .StoreConnection;

System.Data.SqlClient.SqlCommand l_cmd = 
                  new System.Data.SqlClient.SqlCommand(query_arg);
l_cmd.Connection = (System.Data.SqlClient.SqlConnection) l_connection;
System.Data.SqlClient.SqlDataAdapter l_da = 
                 new    System.Data.SqlClient.SqlDataAdapter(l_cmd);
System.Data.DataSet l_ds = new System.Data.DataSet();     
l_da.Fill(l_ds);

CLONE metadata

System.Data.DataTable l_dt = l_ds.Tables[0].Clone();    

back to linq via System.Data.DataSetExtensions

var dt = (from data in l_ds.Tables[0].AsEnumerable()
                  select data).ToList();

   foreach (DataColumn column in l_dt.Columns)
   {
      var binding = new Binding(string.Format("[{0}]", column.Ordinal));
      datagrid.Columns.Add(new DataGridTextColumn() 
                     { Header = column.ColumnName, Binding = binding });
   }

        datagrid.ItemsSource = dt;

Answer 5

If you are using MVVM pattern, I think this is better solution:

Create ObservableCollecion in your ViewModel, which will be binded to ItemSource in DataGrid :

public ObservableCollection<T> bindableCollection;

Then, fetch data from your database, example:

public void RefreshDataGrid(){
    using (var conn = new SqlConnection(ConfigurationManager.ConnectionStrings[
        "RecipeManager.Properties.Settings.RecipeManagerConnectionString"].ConnectionString)){
       var fetchedData = conn.Query<Flavour>("select * from [Flavour]");
        ConvertToDataTable(fetchedData);

    }
}

Last step, and the most important, create function, which will add IEnumerable to your ObservableCollection:

private void ConvertToObservableCollection(IEnumerable<Flavour> items){
    ObservableCollection<Flavour> flavours = new ObservableCollection<Flavour>();
    foreach (var item in items){
        Flavour flavour = item as Flavour;
        flavours.Add(new Flavour(flavour.Name,flavour.Company,flavour.Shop,flavour.Amount));
    }
    Flavours = flavours;
}

I think, that this is good approach for mvvm.