Add Computed Column to a DataTable

Add Computed Column to a DataTable

• Craig Cormier

  Thanks for that. Much appreciated.

  (PS - Why was this thread moved Got nothing to do with Windows Forms..)

  
  

• Charles Darwin

  I have a similar problem like Andrew.

  
  I want to add a Computed Column to a DataTable. This column should be bindable, sortable, ... like any other column. I tried the following:
  

  • I use a typed DataSet (dsData) which is filled with ReadXml and written to the disk with WriteXml. The DataSet contains a Table (dtPerson) with the two Columns Firstname and Lastname. Now I want to Add the Computed Column FullName by extending the DataSet.
  • Extend the dsData-Class:
    
    Partial Class dsData
    Partial Class dtPerson
    
    Private FullNameColumn as DataColumn
    
    Public ReadOnly Property FullName() as DataColumn
    Get
    Return FullNameColumn
    End Get
    End Property
    
    Private Sub dtPersonDataTable_Initialized(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Initialized
    FullNameColumn = New System.Data.DataColumn("FullName", GetType(String), Nothing, System.Data.MappingType.Element)
    MyBase.Columns.Add(FullNameColumn)
    End Sub
    
    Private Sub dtPersonDataTable_ColumnChanged(ByVal sender As Object, ByVal e As System.Data.DataColumnChangeEventArgs) Handles Me.ColumnChanged
    Select Case e.Column.ColumnName
    Case Me.FirstNameColumn.ColumnName
    CType(e.Row, dtPersonRow).Item(FullNameColumn.ColumnName) = CType(e.Row, dtPersonRow).FirstName & " " & CType(e.Row, dtPersonRow).LastName
    Case Me.LastNameColumn.ColumnName
    CType(e.Row, dtPersonRow).Item(FullNameColumn.ColumnName) = CType(e.Row, dtPersonRow).FirstName & " " & CType(e.Row, dtPersonRow).LastName
    End Select
    End Sub
    
    End Class
    End Class

  This works quite good. I can bind this column to a DataGridView, but I have to bind it by code because it doesn't show up as a Column of the DataTable in the designer (DataSources).
  Another Problem is that the column is also written to the XML-File with WriteXml (and therefore the written data isn't consistent to the written Schema).
  So it works, but I'm not really happy with it.

  Does anyone know a better solution to create a computed Column
  (I think in this example I could achieve this by creating a Column with an expression. But I want a solution in which i can write code to compute the value).

  Robert

  
  

• taral

  I've been facing similar problems - I guess I hoped that partial classes in .NET 2.0 would make customising strongly typed datasets a bit easier (maybe .NET 3.0, eh!). A way round this (I guess an answer to (b) above) is to add a handler for OnRowDataBound event of the gridview that you're binding to. So, for your example above:

  (1) Modify your extended typed dataset - you need to add the property to the relevant dataRow class rather than the data table. Something like:-

  Partial Public Class TipperRoundResults
  Partial Public Class stp_GetViewTipsDataRow
  Public ReadOnly Property SomeNewField() As String
  Get
  Return "Yipee!"
  End Get
  End Property
  End Class
  End Class

  (2) Replace the <asp:BoundField ../> column with an <asp:TemplateField../> like this:

  <asp:TemplateField HeaderText="Some New Field"><ItemTemplate><asp:Label ID="lblSomeNewField" runat="server" /></ItemTemplate></asp:TemplateField>

  (3) Add a link to the handler to your gridview:

  <asp:GridView ....

  OnRowDataBound="grdProperties_RowDataBound" />

  (4) Code the handler

  protected void grdProperties_RowDataBound(object sender, GridViewRowEventArgs e)
  {
  // Fired for each row being bound
  if (e.Row.RowType == DataControlRowType.DataRow)
  {
  // Find the control in the template column for this row
  Label oLabel = (Label)e.Row.FindControl("lblSomeNewField");

  oLabel.Text = ((TipperRoundResults.stp_GetViewTipsDataRow) ((System.Data.DataRowView) e.Row.DataItem).Row).SomeNewField;
  }
  }

  hope this saves somebody some messing around!

  }

  
  
• Add Computed Column to a DataTable