SqlErrorCollection.Count SqlErrorCollection.Count SqlErrorCollection.Count SqlErrorCollection.Count Property

Определение

Получает число ошибок в коллекции.Gets the number of errors in the collection.

public:
 property int Count { int get(); };
public int Count { get; }
member this.Count : int
Public ReadOnly Property Count As Integer

Значение свойства

Общее число ошибок в коллекции.The total number of errors in the collection.

Реализации

Примеры

В следующем примере каждый SqlError объект SqlErrorCollection отображается в коллекции.The following example displays each SqlError within the SqlErrorCollection collection.

public static void ShowSqlException(string connectionString)
{
    string queryString = "EXECUTE NonExistantStoredProcedure";
    
    using (SqlConnection connection = new SqlConnection(connectionString))
    {
        SqlCommand command = new SqlCommand(queryString, connection);
        try
        {
            command.Connection.Open();
            command.ExecuteNonQuery();
        }
        catch (SqlException ex)
        {
            DisplaySqlErrors(ex);
        }
    }
}

private static void DisplaySqlErrors(SqlException exception)
{
    for (int i = 0; i < exception.Errors.Count; i++)
    {
        Console.WriteLine("Index #" + i + "\n" +
            "Error: " + exception.Errors[i].ToString() + "\n");
    }
    Console.ReadLine();
}
Public Sub ShowSqlException(ByVal connectionString As String)
    Dim queryString As String = "EXECUTE NonExistantStoredProcedure"

    Using connection As New SqlConnection(connectionString)
        Dim command As New SqlCommand(queryString, connection)

        Try
            command.Connection.Open()
            command.ExecuteNonQuery()

        Catch ex As SqlException
            DisplaySqlErrors(ex)
        End Try
    End Using
End Sub

Private Sub DisplaySqlErrors(ByVal exception As SqlException)
    Dim i As Integer

    For i = 0 To exception.Errors.Count - 1
        Console.WriteLine("Index #" & i & ControlChars.NewLine & _
            "Error: " & exception.Errors(i).ToString() & ControlChars.NewLine)
    Next i
    Console.ReadLine()
End Sub

Применяется к

Дополнительно