避免使用 sum 的笛卡尔积

Question

我想总结 tickets table 中的 stake，按 customer_id 和 bonus 中的 date_trunc('day') 分组 table.

问题是行数在增加，不知道怎么解决

https://www.db-fiddle.com/f/yWCvFamMAY9uGtoZupiAQ/4

CREATE TABLE tickets (
    ticket_id integer,
    customer_id integer,
  stake integer,
  reg_date date
);

CREATE TABLE bonus (
bonus_id integer,
customer_id integer,
reg_date date
);

insert into tickets 
values
(1,100, 12,'2019-01-10 11:00'),
(2,100, 10,'2019-01-10 12:00'),
(3,100, 30,'2019-01-10 13:00'),
(4,100, 10,'2019-01-11 14:00'),
(5,100, 15,'2019-01-11 15:00'),

(6,102, 25,'2019-01-10 10:00'),
(7,102, 25,'2019-01-10 11:10'),
(8,102, 13,'2019-01-11 12:40'),
(9,102, 9,'2019-01-12 15:00'),
(10,102, 7,'2019-01-13 18:00'),


(13,103, 15,'2019-01-12 19:00'),
(14,103, 11,'2019-01-12 22:00'),
(15,103, 11,'2019-01-14 02:00'),
(16,103, 11,'2019-01-14 10:00')
;

insert into bonus
values
(200,100,'2019-01-10 05:00'),
(201,100,'2019-01-10 06:00'),
(202,100,'2019-01-10 15:00'),
(203,100,'2019-01-10 15:50'),
(204,100,'2019-01-10 16:10'),
(205,100,'2019-01-10 16:15'),
(206,100,'2019-01-10 16:22'),
(207,100,'2019-01-11 10:10'),
(208,100,'2019-01-11 16:10'),

(209,102,'2019-01-10 10:00'),
(210,102,'2019-01-10 11:00'),
(211,102,'2019-01-10 12:00'),
(212,102,'2019-01-10 13:00'),

(213,103,'2019-01-11 11:00'),
(214,103,'2019-01-11 18:00'),
(215,103,'2019-01-12 15:00'),
(216,103,'2019-01-12 16:00'),
(217,103,'2019-01-14 02:00')




select 
customer_id, 
date_trunc('day', b.reg_date), 
sum(t.stake)

from tickets t
join bonus b using (customer_id)
where date_trunc('day', b.reg_date) = date_trunc('day', t.reg_date)
group by 1,2
order by 1

客户 102 的输出应为：

102,2019-01-10, 50

Answer 1

好的，我想你想获取 tickets table 列 stake 的汇总数据，并且记录的 customer_id, reg_date 对出现在第二个 tablebonus，所有的业务都与bonus_id无关，我说的对吗？ bonus中的customer_id, reg_date对是重复的，所以你需要一个distinct，然后join来自tickets.The的sum数据完成SQL 结果如下：

with stake_sum as (
select
    customer_id,
    reg_date,
    sum(stake)
from
    tickets
group by
    customer_id,
    reg_date
)
,bonus_date_distinct as (
select
    distinct customer_id,
    reg_date
from
    bonus
)
select
    a.*
from
    stake_sum a
join
    bonus_date_distinct b on a.customer_id = b.customer_id and a.reg_date = b.reg_date order by customer_id, reg_date;
 customer_id |  reg_date  | sum 
-------------+------------+-----
         100 | 2019-01-10 |  52
         100 | 2019-01-11 |  25
         102 | 2019-01-10 |  50
         103 | 2019-01-12 |  26
         103 | 2019-01-14 |  22
(5 rows)

避免使用 sum 的笛卡尔积

Avoid cartesian product using sum

sql

cartesian-product

postgresql-9.4