获取 DATEDIFF 结果的 SUM

Question

我有一个 SQL 语句（MS SQL Server 2012），可以简单计算某些记录中的日期差异。我也想 return DATEDIFF 的总和。

这是我的查询：

with cte as (
    select ROW_NUMBER () OVER (ORDER BY Date) as ID, Die_ID, Date, Status
    from Tooling_Status
    where (date between '2018-02-27 00:00:00' and '2019-02-27 11:59:59')
    and Date is not null)
select DATEDIFF (ss, c1.Date, min (c2.Date)) as Seconds, convert (nvarchar (10), c1.Date, 120) as DIA
from cte c1
left outer join cte c2
on c1.Date < c2.Date and c1.Die_ID = c2.Die_ID
where c1.Status = 2 and c2.Status = 1
group by c1.Date
order by DIA desc

这是我的结果：

Seconds DAY
2253    2019-02-27
166     2019-02-27
17      2019-02-27
104     2019-02-27
36      2019-02-27
11      2019-02-07
20      2019-02-07
32      2019-02-07
22      2019-02-07
27      2019-02-07
33      2019-02-07
15      2019-02-07
34      2019-02-07
120     2019-02-04
99420   2019-01-07
60      2018-09-26

我需要这个结果：

Seconds   DAY
2732      2019-02-27
194       2019-02-07
120       2019-02-04
99420     2019-01-07
60        2018-09-26

感谢帮助！

Answer 1

少量调整

使用派生的额外 GROUP BY table
删除了不必要的字符串转换
正在格式化...

SQL:

WITH cte
AS (SELECT ROW_NUMBER() OVER (ORDER BY Date) AS ID,
           Die_ID,
           Date,
           Status
    FROM Tooling_Status
    WHERE (date      BETWEEN '2018-02-27 00:00:00' AND '2019-02-27 11:59:59'      )
          AND Date IS NOT NULL
)

SELECT SUM(d.Seconds) AS Seconds 
, d.DIA AS [Day]
FROM (
     SELECT DATEDIFF(ss, c1.Date, MIN(c2.Date)) AS Seconds,
           CAST(c1.Date AS DATE) AS DIA
     FROM cte c1
        LEFT OUTER JOIN cte c2
            ON c1.Date < c2.Date
               AND c1.Die_ID = c2.Die_ID
     WHERE c1.Status = 2 AND c2.Status = 1
     GROUP BY c1.Date
) d
GROUP BY d.DIA
ORDER BY [Day] DESC;

获取 DATEDIFF 结果的 SUM

Get the SUM of a DATEDIFF result

sql-server

sum

datediff