Question

我每小时将不同用户钱包的余额存储在数据库中：

Table: snapshots

id      wall_id   balance          created_at
--------------------------------------------------
...

1300    30        1461.18721408    2019-07-05 18:00:05
1301    30        2922.37442816    2019-07-05 19:00:05
1302    30        228.31050220     2019-07-05 20:00:05
1303    30        2283.10502200    2019-07-05 21:00:05
1304    30        285.38812775     2019-07-05 22:00:05
1305    30        57.07762555      2019-07-05 23:00:05 <--
1306    25        2511.41552420    2019-07-05 18:00:05
1307    25        251.14155242     2019-07-05 19:00:05
1308    25        21735.61642968   2019-07-05 20:00:05
1309    25        22523.28766227   2019-07-05 21:00:05
1310    25        79.90867577      2019-07-05 22:00:05
1311    25        285.38812775     2019-07-05 23:00:05 <--
1312    100       2511.41552420    2019-07-05 18:00:05
1313    100       251.14155242     2019-07-05 19:00:05
1314    100       21735.61642968   2019-07-05 20:00:05
1315    100       22523.28766227   2019-07-05 21:00:05
1316    100       79.90867577      2019-07-05 22:00:05
1317    100       285.38812775     2019-07-05 23:00:05
1318    30        1461.18721408    2019-07-06 18:00:05
1319    30        2922.37442816    2019-07-06 19:00:05
1320    30        228.31050220     2019-07-06 20:00:05
1321    30        2283.10502200    2019-07-06 21:00:05
1322    30        285.38812775     2019-07-06 22:00:05
1323    30        79.90867577      2019-07-06 23:00:05 <--
1324    25        2511.41552420    2019-07-06 18:00:05
1325    25        251.14155242     2019-07-06 19:00:05
1326    25        21735.61642968   2019-07-06 20:00:05
1327    25        22523.28766227   2019-07-06 21:00:05
1328    25        79.90867577      2019-07-06 22:00:05
1329    25        21735.61642968   2019-07-06 23:00:05 <--
1330    100       2511.41552420    2019-07-06 18:00:05
1331    100       251.14155242     2019-07-06 19:00:05
1332    100       21735.61642968   2019-07-06 20:00:05
1333    100       22523.28766227   2019-07-06 21:00:05
1334    100       79.90867577      2019-07-06 22:00:05
1335    100       285.38812775     2019-07-06 23:00:05

...

现在我想获取在wall.id = 30 && 25的每一天所做的最后快照的总和。

要了解算法，我标记了应添加的值。

F.e。：

Example result for wall_id 30 & 25 on EOD

day                           sumEOD 
--------------------------------------------------
2019-07-05 00:00:00           ‭342.4657533‬
2019-07-06 00:00:00           21815.52510545

还应该可以对其进行修改，并获取在wall.id = 30 && 25的情况下每周进行的最新快照的总和。

实现这一目标的最简单方法是什么？我正在运行postgresql。

Answer 1

demo:db<>fiddle

SELECT 
    created_date,
    SUM(balance) AS sum_balance
FROM (
    SELECT DISTINCT ON (1, 2)
        wall_id,
        created_at::date AS created_date,
        balance
    FROM
        snapshots
    WHERE wall_id IN (25, 30)
    ORDER BY 1, 2, created_at DESC
) s
GROUP BY created_date

DISTINCT ON为您提供有序组的第一条记录。在这种情况下，组为(wall_id, created_at::date)。 created_at::date给出了时间戳的一部分。当然，顺序是按组排序的，在该组内，记录按其时间戳记DESC进行排序，时间戳记将每天的最新记录排序到组的顶部。这个由DISTINCT ON拍摄。

之后，您可以简单地将结果分组。

Answer 2

首先使用group by wall_id, date(created_at)查找每天制作的最后快照，然后加入表并按天获取总和：

select 
  date(s.created_at) "day",
  sum(s.balance) sumEOD
from snapshots s inner join (
  select wall_id, max(created_at) maxdate
  from snapshots
  where wall_id in ('25', '30')
  group by wall_id, date(created_at)
) g on g.wall_id = s.wall_id and g.maxdate = s.created_at  
group by "day"
order by "day"

请参见demo。
结果：

| day                      | sumeod         |
| ------------------------ | -------------- |
| 2019-07-05T00:00:00.000Z | 342.4657533    |
| 2019-07-06T00:00:00.000Z | 21815.52510545 |

如果您想获取每周的总和，则可以更改：

group by wall_id, date(created_at)

收件人：

group by wall_id, date_trunc('week', created_at)

获取一天/每月/一周结束时的值总和

2 个答案: