我有一个像这样的样本表：
purchase_datetime customer_id value purchase_id
2013-01-08 17:13:29 45236 92 2526
2013-01-03 15:42:35 45236 16 2565
2013-01-03 15:42:35 45236 16 2565
2013-03-08 09:04:52 45236 636 2563
2013-12-08 12:12:24 45236 23 2505
2013-12-08 12:12:24 45236 23 2505
2013-12-08 12:12:24 45236 23 2505
2013-12-08 12:12:24 45236 23 2505
2013-07-08 22:35:53 35536 73 2576
2013-07-08 09:52:03 35536 4 5526
2013-10-08 16:23:29 52626 20 2226
...
2013-04-08 17:49:31 52626 27 4526
2013-12-09 20:40:53 52626 27 4626
现在我需要找到客户在过去几个月内每次购买（purchase_id）时所花费的总金额（价值）。但我有一个问题，因为buy_id加倍，所以我需要在purchase_id做区别。
这是我到目前为止没有明显的，我不知道如何接近分明。
Select customer_id
sum(case when ( date '2017-01-01' - purchase_datetime::DATE <=30) then value else 0 end) as 1month,
sum( case when ( date '2017-01-01' - purchase_datetime::DATE <=90) then value else 0 end) as 3month,
sum( case when ( date '2017-01-01' - purchase_datetime::DATE <=180) then value else 0 end) as 6month,
sum( case when ( date '2017-01-01' - purchase_datetime::DATE <=360) then value else 0 end) as 12month
FROM table_data
GROUP BY (customer_id)
ORDER BY amount_1month DESC;
使用window func可能更好吗？
期望的输出：
purchase_datetime customer_id value purchase_id
2013-01-08 17:13:29 45236 92 2526
2013-01-03 15:42:35 45236 16 2565
2013-03-08 09:04:52 45236 636 2563
2013-12-08 12:12:24 45236 23 2505
2013-07-08 22:35:53 35536 73 2576
2013-07-08 09:52:03 35536 4 5526
2013-10-08 16:23:29 52626 20 2226
...
2013-04-08 17:49:31 52626 27 4526
2013-12-09 20:40:53 52626 27 4626
答案 0 :(得分：1)
您可以选择子查询，并在该子查询中使用DISTINCT（或GROUP BY）。
例如：
SELECT
customer_id,
sum(case when purchase_datetime::DATE between current_date - interval '1 month' and current_date then "value" else 0 end) as "1month",
sum(case when purchase_datetime::DATE between current_date - interval '3 month' and current_date then "value" else 0 end) as "3month",
sum(case when purchase_datetime::DATE between current_date - interval '6 month' and current_date then "value" else 0 end) as "6month",
sum(case when purchase_datetime::DATE between current_date - interval '1 year' and current_date then "value" else 0 end) as "12month"
FROM (
select
distinct purchase_id, customer_id, purchase_datetime, "value"
-- distinct on (purchase_id) customer_id, purchase_datetime, "value"
-- Note: with this type of distinct you assume that for each purchase_id there is only 1 combination of the 3 other field values.
from table_data
) p
GROUP BY customer_id
ORDER BY "1month" DESC;
TESTDATA：
create table table_data (purchase_datetime timestamp(0),customer_id int,"value" int,purchase_id int);
insert into table_data (purchase_datetime,customer_id,"value",purchase_id) values
(current_timestamp - interval '11 month',45236,92,2526),
(current_timestamp - interval '11 month',45236,16,2565),
(current_timestamp - interval '1 month',45236,16,2565),
(current_timestamp - interval '2 month',45236,636,2563),
(current_timestamp - interval '5 month',45236,23,2505),
(current_timestamp - interval '5 month',45236,23,2505),
(current_timestamp - interval '5 month',45236,23,2505),
(current_timestamp - interval '3 month',35536,73,2576),
(current_timestamp - interval '2 month',35536,4,5526),
(current_timestamp - interval '1 month',52626,20,2226),
(current_timestamp - interval '6 month',52626,27,4526),
(current_timestamp - interval '6 month',52626,27,4626);
答案 1 :(得分：0)
select customer_id, sum(value)
from (
select distinct on (purchase_id) *
from t
) s
where purchase_datetime >= '2017-07-01'
group by 1
;
customer_id | sum
-------------+-----
35536 | 77
52626 | 20
45236 | 23