我有从SQL Server出席的这些数据,我必须分成2列(进出)。
这是一个例子:
这是我想要的结果
查询:
CREATE TABLE [dbo].[Table1]
(
[ID] [varchar](50) NULL,
[DATETIME] [datetime] NULL,
[Flag] [int] NULL
)
INSERT INTO table1
VALUES ('ID-1', '2019-03-13 09:48:00.000', '2'),
('ID-1', '2019-03-13 09:48:00.000', '2'),
('ID-1', '2019-03-13 18:11:00.000', '3'),
('ID-1', '2019-03-13 18:11:00.000', '3'),
('ID-1', '2019-03-14 02:00:00.000', '3'),
('ID-1', '2019-03-14 09:54:00.000', '2'),
('ID-1', '2019-03-14 09:54:00.000', '2'),
('ID-1', '2019-03-14 09:54:00.000', '2'),
('ID-1', '2019-03-14 22:00:00.000', '3'),
('ID-1', '2019-03-14 22:00:00.000', '3'),
('ID-1', '2019-03-14 22:00:00.000', '3'),
('ID-1', '2019-03-15 13:55:00.000', '2'),
('ID-1', '2019-03-15 13:55:00.000', '2'),
('ID-1', '2019-03-15 13:55:00.000', '2'),
('ID-1', '2019-03-15 13:55:00.000', '2'),
('ID-1', '2019-03-15 13:55:00.000', '2'),
('ID-1', '2019-03-15 22:00:00.000', '3'),
('ID-1', '2019-03-15 22:00:00.000', '3'),
('ID-1', '2019-03-15 22:00:00.000', '3'),
('ID-1', '2019-03-15 22:00:00.000', '3')
注意:
我怎样才能做到这一点?
也许尝试从Ins到出局的左连接,并在那之后找到最早的时间。
以下样本(未经测试):
WITH ins
AS ( SELECT *
FROM table1
WHERE Flag = 2 )
,outs
AS ( SELECT *
FROM table1
WHERE Flag = 3 )
SELECT i.ID
,i.DATETIME InTime
,MIN(o.DATETIME) OutTime
FROM ins i
LEFT JOIN outs o ON o.ID = i.ID AND o.DATETIME > i.DATETIME -- out must be later than in
GROUP BY i.ID
,i.DATETIME
,i.Flag;