DataTable Konstruktorer

public:
 DataTable();

public DataTable();

Public Sub New ()

Exempel

I följande exempel skapas en ny DataTable med DataColumn och DataRowoch visas i en DataGridView kontroll.

private void MakeDataTableAndDisplay()
{
    // Create new DataTable.
    DataTable table = new DataTable();

    // Declare DataColumn and DataRow variables.
    DataColumn column;
    DataRow row;

    // Create new DataColumn, set DataType, ColumnName
    // and add to DataTable.
    column = new DataColumn();
    column.DataType = System.Type.GetType("System.Int32");
    column.ColumnName = "id";
    table.Columns.Add(column);

    // Create second column.
    column = new DataColumn();
    column.DataType = Type.GetType("System.String");
    column.ColumnName = "item";
    table.Columns.Add(column);

    // Create new DataRow objects and add to DataTable.
    for(int i = 0; i < 10; i++)
    {
        row = table.NewRow();
        row["id"] = i;
        row["item"] = "item " + i;
        table.Rows.Add(row);
    }
    // Set to DataGrid.DataSource property to the table.
    dataGrid1.DataSource = table;
}

Private Sub MakeDataTableAndDisplay()
   ' Create new DataTable.
   Dim table As New DataTable

   ' Declare DataColumn and DataRow variables.
   Dim column As DataColumn
   Dim row As DataRow

   ' Create new DataColumn, set DataType, ColumnName 
   ' and add to DataTable.    
   column = New DataColumn
   column.DataType = System.Type.GetType("System.Int32")
   column.ColumnName = "id"
   table.Columns.Add(column)

   ' Create second column.
   column = New DataColumn
   column.DataType = Type.GetType("System.String")
   column.ColumnName = "item"
   table.Columns.Add(column)

   ' Create new DataRow objects and add to DataTable.    
   Dim i As Integer
   For i = 0 To 10
      row = table.NewRow
      row("id") = i
      row("item") = "item " & i
      table.Rows.Add(row)
   Next i

   ' Set to DataGrid.DataSource property to the table.
   DataGrid1.DataSource = table
End Sub

Kommentarer

Konstruktorn anger initiala värden för alla egenskaper för DataTable objektet. I följande tabell visas egenskaperna och deras standardvärden. När en instans av DataTable skapas anges följande läs-/skrivegenskaper till initiala värden.

Du kan ändra värdet för någon av dessa egenskaper genom ett separat anrop till egenskapen.

Se även

• DataTables

public:
 DataTable(System::String ^ tableName);

public DataTable(string tableName);

new System.Data.DataTable : string -> System.Data.DataTable

Public Sub New (tableName As String)

Parametrar

Exempel

I följande exempel skapas en DataTable och visas i en DataGridView kontroll.

private void MakeDataTableAndDisplay()
{
    // Create new DataTable.
    DataTable table = new DataTable("table");

    // Declare DataColumn and DataRow variables.
    DataColumn column;
    DataRow row;

    // Create new DataColumn, set DataType,
    // ColumnName and add to DataTable.
    column = new DataColumn();
    column.DataType = System.Type.GetType("System.Int32");
    column.ColumnName = "id";
    table.Columns.Add(column);

    // Create second column.
    column = new DataColumn();
    column.DataType = Type.GetType("System.String");
    column.ColumnName = "item";
    table.Columns.Add(column);

    // Create new DataRow objects and add to DataTable.
    for(int i = 0; i < 10; i++)
    {
        row = table.NewRow();
        row["id"] = i;
        row["item"] = "item " + i;
        table.Rows.Add(row);
    }
    // Set to DataGrid.DataSource property to the table.
    dataGrid1.DataSource = table;
}

Private Sub MakeDataTableAndDisplay()
   ' Create new DataTable.
   Dim table As New DataTable("table")

   ' Declare DataColumn and DataRow variables.
   Dim column As DataColumn
   Dim row As DataRow

   ' Create new DataColumn, set DataType, 
   ' ColumnName and add to DataTable.    
   column = New DataColumn
   column.DataType = System.Type.GetType("System.Int32")
   column.ColumnName = "id"
   table.Columns.Add(column)

   ' Create second column.
   column = New DataColumn
   column.DataType = Type.GetType("System.String")
   column.ColumnName = "item"
   table.Columns.Add(column)

   ' Create new DataRow objects and add to DataTable.    
   Dim i As Integer
   For i = 0 To 10
      row = table.NewRow
      row("id") = i
      row("item") = "item " & i
      table.Rows.Add(row)
   Next i

   ' Set to DataGrid.DataSource property to the table.
   DataGrid1.DataSource = table
End Sub

Se även

• DataTables

protected:
 DataTable(System::Runtime::Serialization::SerializationInfo ^ info, System::Runtime::Serialization::StreamingContext context);

protected DataTable(System.Runtime.Serialization.SerializationInfo info, System.Runtime.Serialization.StreamingContext context);

new System.Data.DataTable : System.Runtime.Serialization.SerializationInfo * System.Runtime.Serialization.StreamingContext -> System.Data.DataTable

Protected Sub New (info As SerializationInfo, context As StreamingContext)

Parametrar

Undantag

Kommentarer

Den här implementeringen av DataTable konstruktorn krävs för ISerializable.

Se även

• DataTables

public:
 DataTable(System::String ^ tableName, System::String ^ tableNamespace);

public DataTable(string tableName, string tableNamespace);

new System.Data.DataTable : string * string -> System.Data.DataTable

Public Sub New (tableName As String, tableNamespace As String)

Parametrar

Se även

• DataTables