SELECT [Month] = DATENAME(MONTH, M), Sponsor, Client, c,
[dates] = STUFF((SELECT ', ' + RTRIM(DATEPART(DAY, [date]))
FROM dbo.booking AS b
WHERE b.Sponsor = x.Sponsor
AND b.Client = x.Client
AND b.[date] >= x.M AND b.[date] < DATEADD(MONTH, 1, x.M)
ORDER BY [date]
FOR XML PATH('')), 1, 2, '')
FROM
(
SELECT
M = DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'),
Sponsor,
Client,
COUNT(booking_id) AS c
FROM dbo.booking
GROUP BY DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'),
Sponsor,
Client
) AS x
ORDER BY M, Sponsor, Client;
Perhatikan bahwa jika kombinasi sponsor/klien memiliki dua pemesanan pada hari yang sama, nomor hari akan muncul dalam daftar dua kali.
EDIT Inilah cara saya menguji:
DECLARE @booking TABLE
(
booking_id INT IDENTITY(1,1) PRIMARY KEY,
[date] DATE,
Sponsor VARCHAR(32),
Client VARCHAR(32)
);
INSERT @booking([date], Sponsor, Client) VALUES
('20120312','AB','y'), ('20120315','AB','y'), ('20120318','AB','y'),
('20120316','FE','x'), ('20120319','FE','x'), ('20120321','FE','x'), ('20120320','FE','x'),
('20120404','AB','x'), ('20120408','AB','x');
SELECT [Month] = DATENAME(MONTH, M), Sponsor, Client, c,
[dates] = STUFF((SELECT ', ' + RTRIM(DATEPART(DAY, [date]))
FROM @booking AS b
WHERE b.Sponsor = x.Sponsor
AND b.Client = x.Client
AND b.[date] >= x.M AND b.[date] < DATEADD(MONTH, 1, x.M)
ORDER BY [date]
FOR XML PATH('')), 1, 2, '')
FROM
(
SELECT
M = DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'),
Sponsor,
Client,
COUNT(booking_id) AS c
FROM @booking
GROUP BY DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'),
Sponsor,
Client
) AS x
ORDER BY M, Sponsor, Client;
Hasil:
Month Sponsor Client c dates
------- ------- ------- ------- --------------
March AB y 3 12, 15, 18
March FE x 4 16, 19, 20, 21
April AB x 2 4, 8