让我说我有一张这样的表:
CREATE TABLE [dbo].[Scheduler](
[DayOfWeek] [tinyint] NOT NULL,
[Time] [time](0) NOT NULL,
[Action] [varchar](255) NOT NULL
)
还有一些数据,如:
INSERT INTO Scheduler VALUES (1, '11:00:00', 'Sunday')
INSERT INTO Scheduler VALUES (2, '12:00:00', 'Monday')
INSERT INTO Scheduler VALUES (4, '13:00:00', 'Tuesday')
INSERT INTO Scheduler VALUES (8, '14:00:00', 'Wednesday')
INSERT INTO Scheduler VALUES (16, '15:00:00', 'Thursday')
INSERT INTO Scheduler VALUES (32, '16:00:00', 'Friday')
INSERT INTO Scheduler VALUES (64, '17:00:00', 'Saturday')
INSERT INTO Scheduler VALUES (62, '06:00:00', 'Every business day')
INSERT INTO Scheduler VALUES (127, '08:00:00', 'Every day')
如果DayOfWeek有多个标志,如何在SELECT语句中生成多行?
例如,这一行:
INSERT INTO Scheduler VALUES (62, '06:00:00', 'Every business day')
它将在SELECT语句中以5行表示(每天一个/标志集)
DayOfWeek Time Message
--------- ---------------- ---------------------------
2 06:00:00 Every business day
4 06:00:00 Every business day
8 06:00:00 Every business day
16 06:00:00 Every business day
32 06:00:00 Every business day
使用所有数据运行相同的查询将给我19行。
我不知道怎么能这样做。 我想我可以使用光标来做到这一点,但这是最好的选择吗?
感谢。
答案 0 :(得分:0)
您需要SQL Server的Bitwise Operators。此示例显示如何确定当前值中包含哪些标志。
DECLARE @Mon INT = 1;
DECLARE @Tue INT = 2;
DECLARE @Wed INT = 4;
DECLARE @MonAndTue INT = 3; -- Mon (1) and Tue (2) = 3.
SELECT
@MonAndTue & @Mon, -- Contains Monday, returns Monday.
@MonAndTue & @Tue, -- Contains Tuesday, returns Tuesday.
@MonAndTue & @Wed -- Does not contain Wednesday, returns 0.
在可能的情况下,我建议在SQL Server中避免使用基于位掩码的解决方案。这些在其他语言中非常出色(C让人想到)。但是,当每列包含单个值时,SQL最有效,描述单个项目。当然,你可以结合这些方法。此表设计允许您保留基本2键(非常适合前端),并包含简单的位字段,使后端中的过滤成为一项简单的任务。
CREATE TABLE [Day]
(
DayId INT PRIMARY KEY,
[DayName] VARCHAR(9),
IsMonday BIT,
IsTuesday BIT,
...
IsSunday BIT
)
;
修改强>
抱歉!我实际上没有回答你的问题。要在连接中使用按位运算,需要使用这些行:
WITH Scheduler AS
(
/* Sample data.
*/
SELECT
*
FROM
(
VALUES
(1, 'Sunday'),
(2, 'Monday'),
(4, 'Tuesday'),
(8, 'Wednesday'),
(16, 'Thursday'),
(32, 'Friday'),
(64, 'Saturday'),
(62, 'Every business day'),
(127, 'Every day')
) AS r(DayId, [DayName])
)
/* This join returns every day from s2 that
* is contained within the s1 record Every business day.
*/
SELECT
*
FROM
Scheduler AS s1
INNER JOIN Scheduler AS s2 ON (s1.DayId & s2.DayId) = s2.DayId
WHERE
s1.DayId = 62
;
此处过滤S1以返回每个工作日。 S2在S1上加入,找到匹配项。这将返回周一,周二,周三等,而不返回星期六和太阳。