pdo*_*dog 6 sql t-sql sql-server sql-server-2008 gaps-and-islands
我已经在这工作了一段时间.我想知道如何得到这张桌子:
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
Run Code Online (Sandbox Code Playgroud)
看起来像这张桌子:
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
Run Code Online (Sandbox Code Playgroud)
我尝试使用游标,但没有必要.此外,它必须分组.连续几天打破一个新的分组发生.任何帮助,将不胜感激.
此查询将生成上述示例数据
;
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
Run Code Online (Sandbox Code Playgroud)
;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
Run Code Online (Sandbox Code Playgroud)
绝对比 @Martin 的解决方案逻辑更加复杂,但我至少应该得到一点,因为他使用了我的 @x 表 - 所以他的解决方案看起来更整洁。:-)
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;
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
1247 次 |
| 最近记录: |