Setup, jadi kami yakin kami membicarakan hal yang sama:
USE tempdb;
GO
CREATE TABLE dbo.users
(
[user_id] INT IDENTITY(1,1) PRIMARY KEY,
hired_date DATE NOT NULL,
termination_date DATE
);
CREATE TABLE dbo.[date table]
(
week_start DATE NOT NULL UNIQUE,
week_end AS CONVERT(DATE, DATEADD(DAY, 6, week_start))
);
GO
SET NOCOUNT ON;
GO
INSERT dbo.[date table](week_start) VALUES
('20110806'),
('20110813'),
('20110820');
INSERT dbo.users(hired_date, termination_date) VALUES
('20110101', NULL), -- long-time, active
('20110101', '20110807'), -- long-time, fired in week 1
('20110807', '20110815'), -- hired week 1, fired week 2
('20110816', '20110816'), -- hired week 2, fired week 2
('20110807', '20110825'), -- hired week 1, fired week 3
('20110806', NULL), -- hired week 1, active
('20110807', NULL), -- hired week 1, active
('20110813', NULL), -- hired week 2, active
('20110821', NULL); -- hired week 3, active
GO
Dengan logika ini, seharusnya ada 6 karyawan aktif selama minggu 1, 7 karyawan aktif selama minggu 2, dan kembali ke 6 lagi di minggu 3. Butuh beberapa menit dan menggambar garis aktif di selembar kertas untuk mencari tahu di mana Saya salah dalam permintaan saya. Sekarang mari kita coba yang ini terhadap data sampel yang telah kita siapkan di tempdb:
;WITH last_8_weeks AS
(
SELECT TOP (8) week_start, week_end
FROM dbo.[date table]
WHERE week_start >= DATEADD(WEEK, -9, CURRENT_TIMESTAMP)
ORDER BY week_start DESC
)
SELECT d.week_end, COUNT(u.user_id)
FROM last_8_weeks AS d
LEFT OUTER JOIN dbo.users AS u
ON u.hired_date <= d.week_end
AND COALESCE(u.termination_date, DATEADD(DAY, 1, d.week_end)) >= d.week_start
GROUP BY d.week_end
ORDER BY d.week_end;
Dan kemudian bersihkan:
GO
DROP TABLE dbo.[date table], dbo.users;