[英]SQL Server 2008, temp tables, cursor
我已經為此工作了一段時間。 我想知道如何獲得這張桌子:
id open_dt ops_hrs
1 10/31/2011 7:00AM - 5:30PM
2 11/1/2011 7:00AM - 5:00PM
3 11/2/2011 7:00AM - 5:00PM
4 11/3/2011 7:00AM - 5:00PM
5 11/6/2011 7:00AM - 7:00PM
6 11/8/2011 7:00AM - 5:00PM
看起來像這張表:
max_date min_date ops_hrs
10/31/2011 10/31/2011 7:00AM - 5:30PM
11/1/2011 11/3/2011 7:00AM - 5:00PM
11/6/2011 11/6/2011 7:00AM - 7:00PM
11/8/2011 11/8/2011 7:00AM - 5:00PM
我嘗試使用 cursor 但沒有必要。 此外,它必須被分組。 只要連續幾天休息,就會發生新的分組。 任何幫助,將不勝感激。
此查詢將生成上述示例數據
;
WITH pdog (id, open_dt,ops_hrs) AS
(
SELECT 1, CAST('10/31/2011' AS datetime), '7:00AM - 5:30PM'
UNION ALL SELECT 2, CAST('11/1/2011' AS datetime),'7:00AM - 5:00PM'
UNION ALL SELECT 3, CAST('11/2/2011' AS datetime),'7:00AM - 5:00PM'
UNION ALL SELECT 4, CAST('11/3/2011' AS datetime),'7:00AM - 5:00PM'
UNION ALL SELECT 5, CAST('11/6/2011' AS datetime),'7:00AM - 7:00PM'
UNION ALL SELECT 6, CAST('11/8/2011' AS datetime),'7:00AM - 5:00PM'
)
SELECT * FROM pdog
;WITH CTE
AS ( SELECT * ,
DATEDIFF(DAY, 0, open_dt) - ROW_NUMBER() OVER
( PARTITION BY ops_hrs ORDER BY open_dt ) AS Grp
FROM @x
)
SELECT
MIN(open_dt) AS min_date ,
MAX(open_dt) AS max_date ,
ops_hrs
FROM CTE
GROUP BY ops_hrs ,
Grp
ORDER BY min_date
絕對比@Martin 的解決方案更復雜的邏輯,但我至少應該得到一點,因為他使用了我的@x 表——所以他的解決方案看起來更整潔。 :-)
DECLARE @x TABLE(id INT IDENTITY(1,1), open_dt DATE, ops_hrs VARCHAR(32));
INSERT @x(open_dt, ops_hrs) VALUES
('2011-10-31', '7:00AM - 5:30PM'),
('2011-11-01', '7:00AM - 5:00PM'),
('2011-11-02', '7:00AM - 5:00PM'),
('2011-11-03', '7:00AM - 5:00PM'),
('2011-11-06', '7:00AM - 7:00PM'),
('2011-11-08', '7:00AM - 5:00PM');
;WITH d AS
(
SELECT open_dt, ops_hrs, max_date = COALESCE((SELECT MAX(open_dt)
FROM @x AS b WHERE b.open_dt > a.open_dt
AND NOT EXISTS (SELECT 1 FROM @x AS c
WHERE c.open_dt >= a.open_dt
AND c.open_dt < b.open_dt
AND c.ops_hrs <> b.ops_hrs)), open_dt)
FROM @x AS a
)
SELECT
min_date = MIN(open_dt),
max_date,
ops_hrs
FROM d
GROUP BY max_date, ops_hrs
ORDER BY min_date;
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