Get the sum of turnover in the last row SQL Server 2008 - sql-server

Question

Solution: I had to add a WHERE clause to both SELECT queries.
I'm working on a SQL Server 2008 database.
I'm trying to write a SQL query to return a resultset like that :
date | turnover_centrale | turnover_public
+------------+----------------------+--------------------
| 2017-02-14 | 233.34 | 383.83
| 2017-03-14 | 142.81 | 166.8
| 2017-04-14 | 173.25 | 250.51
| 2017-05-14 | 186.96 | 245.08
| 2017-06-14 | 61.26 | 97.67
| 2017-07-14 | 262.98 | 356.16
| 2017-08-14 | 89.88 | 162.38
| 2017-09-14 | 250.32 | 381.47
| 2017-10-14 | 386.06 | 581.96
| Total | Result of the column | Result of the column
My SQL query is :
SELECT
CLC_DATE, CLC_PRIX_CENTRALE, CLC_PRIX_PUBLIC
FROM
##.dbo.CLIENTS_CONSO
WHERE
CF_USER = :ref
UNION ALL
SELECT
NULL, SUM(CLC_PRIX_CENTRALE), SUM(CLC_PRIX_PUBLIC)
FROM
##.dbo.CLIENTS_CONSO
The result of the last row is unfortunately false :(
So do you have a better solution ?
Thanks a lot!

Answer 1

Considering that each date only has one entry, perhaps:
WITH VTE AS (
SELECT CONVERT(date,[date]) AS [date], turnover_centrale, turnover_public
FROM (VALUES
('20170214',233.34,383.83),
('20170314',142.81,166.8 ),
('20170414',173.25,250.51),
('20170514',186.96,245.08),
('20170614',61.26 ,97.67 ),
('20170714',262.98,356.16),
('20170814',89.88 ,162.38),
('20170914',250.32,381.47),
('20171014',386.06,581.96)) V([date],turnover_centrale, turnover_public))
SELECT ISNULL(CONVERT(varchar(10),[date],120),'Total') AS [date],
SUM(turnover_centrale) AS turnover_centrale,
SUM(turnover_public) AS turnover_public
FROM VTE
GROUP BY [date] WITH ROLLUP;

Answer 2

Thanks a lot !
My table is build like that :
https://i.imgur.com/csXl8vl.png
<br>
CLC_DATE = the date of consumption, it's corresponding to each month in a year.
<br>
CLC_PRIX_PUBLIC = the turnover public
<br>
CLC_PRIX_CENTRALE = the turnover centrale
The WHERE clause is important, beacause its the gateway from the client and his consumnption
With your solution, it doesn't work.