Question

我想编写sql查询，它将给出如下的put 查询：

DivideTime('10','10:00:00','11:00:00')

输出：

10:10

10:20

10:30

10:40

10:50

11:00

Answer 1

这可以使用递归公用表表达式

来完成

CREATE FUNCTION [dbo].[get_time_range] (@interval INT, @start_time TIME, @end_time TIME)
RETURNS @tbl TABLE ([time] TIME)
AS
BEGIN
    WITH [cte_recursion] AS
        (
            SELECT DATEADD(MINUTE, @interval, @start_time) [time]

            UNION ALL

            SELECT
                DATEADD(MINUTE, @interval, [time]) [time]
            FROM [cte_recursion]
            WHERE   [time] < @end_time
        )
    INSERT INTO @tbl ([time])
    SELECT
        [time]
    FROM [cte_recursion]
    OPTION (MAXRECURSION 0); -- allows more than 100 iterations

    RETURN;
END
GO

SELECT
    *
FROM [dbo].[get_time_range](10, '10:00:00', '11:00:00');

Answer 2

使用numbers表：

select dateadd(minute, 10*n, @start_time)
from numbers n
where n <= datediff(minute, @start_time, @end_time)/@interval

http://dataeducation.com/you-require-a-numbers-table/

如何在给定时间范围内以特定间隔分解时隙

2 个答案: