好的,我有一个表,其中每个 user_id
都有 operation_timestamp
。要检查每个 user_id
执行操作需要多长时间,我使用查询 max(operation_timestamp)-min(operation_timestamp)
,然后将其转换为小数,假设我得到的结果是:
SHIFT USER ID MIN MAX MAX-MIN DECIMAL
shift1 user_1 08:19:42 09:55:20 01:35:37 1.59
shift2 user_2 10:04:27 10:28:22 00:23:54 0.40
shift2 user_3 10:44:07 10:55:58 00:01:51 0.04
shift2 user_4 06:25:33 10:51:52 04:26:19 4.44
现在我的问题是:如何计算出小数点后的时间总和,以便获得所有用户在事件上花费的总时间?如下所示:
SHIFT TOTAL_DECIMAL
shift1 1.59
shift2 4.88
我尝试过相同的查询,但没有 group by user_id
函数,但它随后计算最大值和最小值而不查看单独的 user_id,所以假设它将计算 shift2
总计如 06:25:33 (user_4) 到 10:55:58 (user_3)
,这会得到小数点后 4.45
的结果。
这是我使用但没有成功的查询:
select
case
when SUBSTR(a.operation_ts,12,13) between '00:00:00.000' and '05:59:59.000' then 'Nights'
when SUBSTR(a.operation_ts,12,13) between '06:00:00.000' and '13:59:59.000' then 'Days'
when SUBSTR(a.operation_ts,12,13) between '14:00:00.000' and '21:59:59.000' then 'Lates'
when SUBSTR(a.operation_ts,12,13) between '22:00:00.000' and '23:59:59.000' then 'Nights'
else 'other'
end as shift,
a.userid,
substr(min(a.operation_ts),11,9),
substr(max(a.operation_ts),11,9),
substr((max(a.operation_ts)-min(a.operation_ts)),10,9) as time_on_go,
round(((substr((max(a.operation_ts)-min(a.operation_ts)),11,2))*3600+(substr((max(a.operation_ts)-min(a.operation_ts)),14,2))*60+(substr((max(a.operation_ts)-min(a.operation_ts)),17,2)))/3600,2) time_decimal
from dc_sys_common:user_operation a
where a.activity_code = 1012
and date(a.operation_ts) between today and today+1
and SUBSTR(a.operation_ts,12,9) between '00:00:00' and '21:59:59'
group by userid, shift
having max(a.operation_ts)-min(a.operation_ts)>'0 00:00:01.000'
order by shift asc
最佳答案
在嵌套的第一个查询上进行 GROUP BY SHIFT:
SELECT `SHIFT`, SUM(`DECIMAL`)
FROM
(
-- your query here
)
GROUP BY `SHIFT`
关于mysql - 每个user_id的时间总和,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42202509/