Saya pikir ini bisa ditangani sebagai celah dan pulau masalah. Perhatikan data input berikut:(sama seperti data sampel OP ditambah dua baris tambahan)
id company_id status_id effective_date
-------------------------------------------
1 10 1 2016-12-15
2 10 1 2016-12-30
3 10 5 2017-02-04
4 10 4 2017-02-08
5 11 5 2017-06-05
6 11 1 2018-04-30
Anda dapat menggunakan kueri berikut:
SELECT t.id, t.company_id, t.status_id, t.effective_date, x.cnt
FROM company_status_history AS t
OUTER APPLY
(
SELECT COUNT(*) AS cnt
FROM company_status_history AS c
WHERE c.status_id = 1
AND c.company_id = t.company_id
AND c.effective_date < t.effective_date
) AS x
ORDER BY company_id, effective_date
untuk mendapatkan:
id company_id status_id effective_date grp
-----------------------------------------------
1 10 1 2016-12-15 0
2 10 1 2016-12-30 1
3 10 5 2017-02-04 2
4 10 4 2017-02-08 2
5 11 5 2017-06-05 0
6 11 1 2018-04-30 0
Sekarang Anda dapat mengidentifikasi status = 1 pulau menggunakan:
;WITH CTE AS
(
SELECT t.id, t.company_id, t.status_id, t.effective_date, x.cnt
FROM company_status_history AS t
OUTER APPLY
(
SELECT COUNT(*) AS cnt
FROM company_status_history AS c
WHERE c.status_id = 1
AND c.company_id = t.company_id
AND c.effective_date < t.effective_date
) AS x
)
SELECT id, company_id, status_id, effective_date,
ROW_NUMBER() OVER (PARTITION BY company_id ORDER BY effective_date) -
cnt AS grp
FROM CTE
Keluaran:
id company_id status_id effective_date grp
-----------------------------------------------
1 10 1 2016-12-15 1
2 10 1 2016-12-30 1
3 10 5 2017-02-04 1
4 10 4 2017-02-08 2
5 11 5 2017-06-05 1
6 11 1 2018-04-30 2
Bidang hitung grp akan membantu kami mengidentifikasi pulau-pulau itu:
;WITH CTE AS
(
SELECT t.id, t.company_id, t.status_id, t.effective_date, x.cnt
FROM company_status_history AS t
OUTER APPLY
(
SELECT COUNT(*) AS cnt
FROM company_status_history AS c
WHERE c.status_id = 1
AND c.company_id = t.company_id
AND c.effective_date < t.effective_date
) AS x
), CTE2 AS
(
SELECT id, company_id, status_id, effective_date,
ROW_NUMBER() OVER (PARTITION BY company_id ORDER BY effective_date) -
cnt AS grp
FROM CTE
)
SELECT company_id,
MIN(effective_date) AS start_date,
CASE
WHEN COUNT(*) > 1 THEN DATEADD(DAY, -1, MAX(effective_date))
ELSE MIN(effective_date)
END AS end_date
FROM CTE2
GROUP BY company_id, grp
HAVING COUNT(CASE WHEN status_id = 1 THEN 1 END) > 0
Keluaran:
company_id start_date end_date
-----------------------------------
10 2016-12-15 2017-02-03
11 2018-04-30 2018-04-30
Yang ingin Anda ketahui hanyalah catatan dari atas yang tumpang tindih dengan interval yang ditentukan.
Demo di sini dengan kasus penggunaan yang agak lebih rumit.
