如何在DataTable中获取Distinct行

[netpot]

DataTable中，某一Column（A)中的记录有重复值，但在适配器填充该DataTable中时候不应该使用distinct，因为另外一关键字Column（B)字段存在多个记录对应该Column（A)。
但有的应用中要需要获取Column（A)的distinct记录，在Datatable中的select方法中无法实现，请指教！

[yarshray]

dataview
private void DemonstrateDataView(){
// Create one DataTable with one column.
DataTable myTable = new DataTable("myTable");
DataColumn colItem = new DataColumn("item",Type.GetType("System.String"));
myTable.Columns.Add(colItem);
// Add five items.
DataRow NewRow;
for(int i = 0; i <5; i++){
NewRow = myTable.NewRow();
NewRow["item"] = "Item " + i;
myTable.Rows.Add(NewRow);
}
// Change the values in the table.
myTable.Rows[0]["item"]="cat";
myTable.Rows[1]["item"] = "dog";
myTable.AcceptChanges();

// Create two DataView objects with the same table.
DataView firstView = new DataView(myTable);
DataView secondView = new DataView(myTable);

// Print current table values.
PrintTableOrView(myTable,"Current Values in Table");

// Set first DataView to show only modified versions of original rows.
firstView.RowStateFilter=DataViewRowState.ModifiedOriginal ;
// Print values.
PrintTableOrView(firstView,"First DataView: ModifiedOriginal");
// Add one New row to the second view.
DataRowView myDataRowView;
myDataRowView=secondView.AddNew();
myDataRowView["item"] = "fish";
// Set second DataView to show modified versions of current rows, or New rows.
secondView.RowStateFilter=DataViewRowState.ModifiedCurrent | DataViewRowState.Added;
// Print modified and Added rows.
PrintTableOrView(secondView, "Second DataView: ModifiedCurrent | Added");
}

private void PrintTableOrView(DataTable t, string label){
// This function prints values in the table or DataView.
Console.WriteLine("\n" + label);
for(int i = 0; i<t.Rows.Count;i++){
Console.WriteLine("\t" + t.Rows[i]["item"]);
}
Console.WriteLine();
}

private void PrintTableOrView(DataView dv, string label){

// This overload prints values in the table or DataView.
Console.WriteLine("\n" + label);
for(int i = 0; i<dv.Count;i++){
Console.WriteLine("\t" + dv[i]["item"]);
}
Console.WriteLine();
}