SQL postgres 聚合/数据透视(按周计)

Question

我有一个 EventLogs 表，它记录了给定事件的详细信息，例如事件日期和费用。

+------+----------+---------------------------+-------------------+
| id   | place_id | start_at                  | total_fee_pennies |
+------+----------+---------------------------+-------------------+
| 4242 | 40       | 2013-10-20 19:00:00 +0100 | 8700              |
| 4288 | 48       | 2013-10-22 20:00:00 +0100 | 8000              |
| 4228 | 141      | 2013-10-17 19:30:00 +0100 | 20000             |
| 4232 | 19       | 2013-10-20 19:30:00 +0100 | 8000              |
| 4239 | 5        | 2013-10-20 19:30:00 +0100 | 6800              |
| 4269 | 6        | 2013-10-20 20:00:00 +0100 | 7000              |
| 4234 | 98       | 2013-10-20 20:00:00 +0100 | 6900              |

我希望能够按周汇总此数据总费用，我相信这是一个 PIVOT？

所以我会为给定的月份选择它们:

"SELECT \"event_logs\".* FROM \"event_logs\"  WHERE (event_logs.start_at BETWEEN '2013-10-01' AND '2013-10-31')"

然后以某种方式通过不同的 place_id 和使用 start_at 的周(通常是一个月 5 周？)以及每周的总费用来汇总它们。

place_id，第 1 周，第 2 周，...

但我不确定该怎么做？

Answer 1

在这里你可以找到how to extract the week number .然后在 CASE 语句中使用周数

SQLFiddle demo

WITH T AS
(
SELECT
EventLogs.*
,
extract(week from start_at) - 
extract(week from date_trunc('month', start_at)) + 1 as WeekNo


 FROM EventLogs  
WHERE (start_at BETWEEN '2013-10-01' AND '2013-10-31')
)

SELECT
place_id,
SUM(CASE WHEN WeekNo=1 THEN total_fee_pennies ELSE 0 END) as Week_1,
SUM(CASE WHEN WeekNo=2 THEN total_fee_pennies ELSE 0 END) as Week_2,
SUM(CASE WHEN WeekNo=3 THEN total_fee_pennies ELSE 0 END) as Week_3,
SUM(CASE WHEN WeekNo=4 THEN total_fee_pennies ELSE 0 END) as Week_4,
SUM(CASE WHEN WeekNo=5 THEN total_fee_pennies ELSE 0 END) as Week_5

from T

GROUP BY place_id