Question

如何从第二天早上8点到早上8点获取每日计数器读数总和？

create table Tbl (counterID int, `timestamp` datetime, `counter` int);
insert into Tbl values
(1, '2017-09-18 07:07:03', 22);
insert into Tbl values
        (1, '2017-09-18 08:08:03',     25);
insert into Tbl values
        (1, '2017-09-18 19:08:33',     30);
insert into Tbl values
        (1, '2017-09-19 10:14:54',      8);
insert into Tbl values
        (1, '2017-09-19 10:15:24',     13);
insert into Tbl values
        (1, '2017-10-20 04:21:08',      5);
insert into Tbl values
        (1, '2017-10-23 14:21:38',     24);
insert into Tbl values
        (1, '2017-10-23 14:22:08',     72);
insert into Tbl values
        (1, '2017-10-23 14:22:38',     86);
insert into Tbl values
        (1, '2017-10-24 03:23:09',    100);
insert into Tbl values
        (1, '2017-10-24 04:23:38',    120);
insert into Tbl values
        (1, '2017-10-24 04:24:08',    125);
insert into Tbl values
        (1, '2017-10-25 14:56:52',      2);
insert into Tbl values
        (1, '2017-10-25 14:57:22',      8);
insert into Tbl values
        (1, '2017-10-25 16:39:22',     21);
insert into Tbl values
        (1, '2017-10-25 16:41:52',     22);
insert into Tbl values
        (1, '2017-10-25 16:42:22',     23);
insert into Tbl values
        (1, '2017-10-25 17:18:13',     26);
insert into Tbl values
        (1, '2017-10-25 17:21:15',     17);
insert into Tbl values
        (1, '2017-10-25 17:21:46',     19);
insert into Tbl values
        (1, '2017-10-25 17:22:46',     41);
insert into Tbl values
        (1, '2017-10-26 08:41:58',      2);
insert into Tbl values
        (1, '2017-10-26 14:02:28',      5);
insert into Tbl values
        (1, '2017-10-30 13:39:20',      1);
insert into Tbl values
        (1, '2017-10-30 13:40:19',      4);

Answer 1

以下是您的查询的解决方案，请查看：

select counterid,date(ts2) ts,sum(counter) counter_sum
from (select counterid,counter,convert_tz(ts,'+08:00','+00:00') ts2
        from tx3) s
group by counterid,date(ts2);

输出

+-----------+------------+-------------+
| counterid | ts         | counter_sum |
+-----------+------------+-------------+
|         1 | 2017-09-17 |          22 |
|         1 | 2017-09-18 |          55 |
|         1 | 2017-09-19 |          21 |
|         1 | 2017-10-19 |           5 |
|         1 | 2017-10-23 |         527 |
|         1 | 2017-10-25 |         179 |
|         1 | 2017-10-26 |           7 |
|         1 | 2017-10-30 |           5 |
+-----------+------------+-------------+

Answer 2

尝试此查询：

select counterid, start, end, sum(counter) from (
    select counterid,
           case when cast(timestamp as time) < '08:00:00' then date_add(cast(`timestamp` as date), interval -1 day)
                else cast(`timestamp` as date) end `start`,
           case when cast(timestamp as time) < '08:00:00' then cast(`timestamp` as date)
                else date_add(cast(`timestamp` as date), interval 1 day) end `end`,
           counter
    from tbl
) `a` group by counterid, start, end

MySQL每日数据从特定时间开始

2 个答案: