Question

我有带时间的Sql列。如果彼此相邻的两列之间的时间差大于1小时，我想输出它们

none

我的输出应该是这样

.box002

是否可以在Sql中编写这样的查询？如果可能，我应该从SQL了解什么来执行该查询？（我知道SQL的基本水平，正在学习中）

谢谢。

Answer 1

您可以像这样使用CTE和ROW_NUMBER：

CREATE TABLE #Dates(DateColumn TIME)
INSERT INTO #Dates(DateColumn) VALUES
    ('04:05:44'),
    ('06:07:48'),
    ('08:15:44'),
    ('09:05:22'),
    ('11:15:27'),
    ('11:35:18'),
    ('12:16:54');*/

WITH CTE AS
(
    SELECT *, ROW_NUMBER() OVER (ORDER BY DateColumn) AS RowNumb
    FROM #Dates AS D
)
SELECT *
FROM CTE AS C
LEFT JOIN CTE AS C2 ON C.RowNumb = C2.RowNumb - 1
WHERE DATEDIFF(MINUTE, C.DateColumn, C2.DateColumn) >= 60

Answer 2

对于MSSQL 2008，您可以使用answer provided by www.Studiosql.nl

下面的代码在MSSQL Server 2012中可以根据需要显示结果。

您可以为此使用LAG TSQL function。

CREATE TABLE #time (timeValue TIME)

INSERT INTO #time 
values
     ('04:05:44'),
     ('06:07:48'),
     ('08:15:44'),
     ('09:05:22'),
     ('11:15:27'),
     ('11:35:18'),
     ('12:16:54');

SELECT temp.*
from
(SELECT timevalue, case when datediff(HH, lag(timevalue,1,timevalue) OVER(order by TIMEVALUE),timevalue) > 1 THEN CONCAT(lag(timevalue,1,timevalue) OVER(order by TIMEVALUE),' - ',timevalue) else null END AS displayvalue 
FROM #time) as temp
where temp.displayValue is not null

SQl比较时间彼此相邻的所有行的差异

2 个答案: