我有一个包含 3 列的数据集：
| Item_id | Sourced_from | Cost |
|---|---|---|
| 1 | Local | 15 |
| 2 | Local | 10 |
| 3 | Local | 20 |
| 4 | International | 60 |
我正在尝试在 PostgreSQL 中编写一个查询来获取本地和国际项目的总数，客户可以在现金限额内购买。对于现金限额 50，这是我期望的输出：
| Local | International |
|---|---|
| 3 | 0 |
我对 PostgreSQL 有相当基本的了解，在谷歌搜索后似乎可以通过递归 CTE 解决这个问题，但我无法弄清楚在这种情况下我应该如何选择我的源种子/锚点。
任何想法，我应该如何处理？
【问题讨论】：
不使用递归 CTE，但仍然有效：
DDL/DML：
create table T
(
id integer primary key generated by default AS IDENTITY,
kind text not null,
cost integer not null
);
insert into T(kind, cost)
values ('local', 15),
('local', 10),
('local', 20),
('international', 60);
-- 4. This outer CTE and the following self-join is only necessary in order to display the rows that have a count() of 0
with sub as
(
-- 3. find the total cost of buying this row + all previous rows, grouped by its kind
select X.kind, sum(X.cost) as cost, X.rn
from (
with cte as (
-- 1. assign an increasing row number on each row from the table ordered by its cost
select *, row_number() over (order by T.cost asc, T.kind) as rn
from T
)
-- 2. self-join the CTE on each row with the same kind, but join it only with the rows that have a row number less than or equal to the current row number
select A.id, A.kind, A.cost, B.rn
from cte as A
join cte as B on A.kind = B.kind and A.rn <= B.rn
) as X
group by X.kind, X.rn
)
select M.kind, count(N.*)
from sub as M -- 5. count only the amount of goods that fit in out budget (i.e. 50)
left outer join sub as N on M.rn = N.rn and N.cost <= 50
group by M.kind
;
输出（db-fiddle）：
+-------------+-----+
|kind |count|
+-------------+-----+
|local |3 |
|international|0 |
+-------------+-----+
【讨论】：
我做了一个CTE例子来解决这个问题：
用
重新创建您的案例create table kp (item_id int, sourced_from varchar, cost int);
insert into kp values (1,'local',15);
insert into kp values (2,'local',10);
insert into kp values (3,'local',20);
insert into kp values (4,'international',60);
以下查询会：
kp 中选择cost 小于50 的项目
list_of_items 中添加item_id
递归位：kp 连接，检查source_from 是否相同且kp.item_id 尚未包含在list_of_items 中（避免多次放置同一项目）total_cost)item_id 添加到list_of_items
WITH RECURSIVE items (item_id, next_item_id, sourced_from, total_cost, nr_items, list_of_items) AS (
SELECT
item_id,
item_id as next_item_id,
sourced_from,
cost as total_cost,
1 as nr_items,
ARRAY[item_id] list_of_items
from kp where cost < 50
UNION ALL
SELECT
kp.item_id,
items.item_id as next_item_id,
items.sourced_from,
items.total_cost + kp.cost total_cost,
items.nr_items + 1 as nr_items,
items.list_of_items || kp.item_id as list_of_items
FROM kp join items
on items.sourced_from=kp.sourced_from
and items.list_of_items::int[] @> ARRAY[kp.item_id] = false
WHERE kp.cost + items.total_cost < 50
)
SELECT * FROM items;
如果您针对上述数据集运行，您最终会得到详细的结果
item_id | next_item_id | sourced_from | total_cost | nr_items | list_of_items
---------+--------------+--------------+------------+----------+---------------
1 | 1 | local | 15 | 1 | {1}
2 | 2 | local | 10 | 1 | {2}
3 | 3 | local | 20 | 1 | {3}
1 | 2 | local | 25 | 2 | {2,1}
1 | 3 | local | 35 | 2 | {3,1}
2 | 1 | local | 25 | 2 | {1,2}
2 | 3 | local | 30 | 2 | {3,2}
3 | 1 | local | 35 | 2 | {1,3}
3 | 2 | local | 30 | 2 | {2,3}
1 | 2 | local | 45 | 3 | {3,2,1}
1 | 3 | local | 45 | 3 | {2,3,1}
2 | 1 | local | 45 | 3 | {3,1,2}
2 | 3 | local | 45 | 3 | {1,3,2}
3 | 1 | local | 45 | 3 | {2,1,3}
3 | 2 | local | 45 | 3 | {1,2,3}
(15 rows)
它显示了 3 个local 项的所有排列。
现在，如果您将最后一个 SELECT 部分替换为
SELECT * FROM items order by nr_items desc, total_cost desc, list_of_items asc limit 1;
您还可以选择项目数量最多且成本最接近预算的组合（我还添加了一个基于list_of_items 的升序排序，以便在多个组合的情况下始终收到相同的结果)，在上述情况下会导致
item_id | next_item_id | sourced_from | total_cost | nr_items | list_of_items
---------+--------------+--------------+------------+----------+---------------
3 | 2 | local | 45 | 3 | {1,2,3}
(1 row)
如果您只对sourced_from 的最大值感兴趣，那么最后一个SELECT 变为
select sourced_from, max(nr_items) nr_items from items group by sourced_from;
预期结果是
sourced_from | nr_items
--------------+----------
local | 3
(1 row)
编辑：为了加快查询速度并避免相同对象的多个排列（例如{1,2,3} 和{1,2,3}），我们可以强制下一个item_id 大于当前一。完整查询
WITH RECURSIVE items (item_id, next_item_id, sourced_from, total_cost, nr_items, list_of_items) AS (
SELECT
item_id,
item_id as next_item_id,
sourced_from,
cost as total_cost,
1 as nr_items,
ARRAY[item_id] list_of_items
from kp where cost < 50
UNION ALL
SELECT
kp.item_id,
items.item_id as next_item_id,
items.sourced_from,
items.total_cost + kp.cost total_cost,
items.nr_items + 1 as nr_items,
items.list_of_items || kp.item_id as list_of_items
FROM kp join items
on items.sourced_from=kp.sourced_from
and items.list_of_items::int[] @> ARRAY[kp.item_id] = false
and items.item_id < kp.item_id
WHERE kp.cost + items.total_cost < 50
)
select * from items;
结果
item_id | next_item_id | sourced_from | total_cost | nr_items | list_of_items
---------+--------------+--------------+------------+----------+---------------
1 | 1 | local | 15 | 1 | {1}
2 | 2 | local | 10 | 1 | {2}
3 | 3 | local | 20 | 1 | {3}
2 | 1 | local | 25 | 2 | {1,2}
3 | 1 | local | 35 | 2 | {1,3}
3 | 2 | local | 30 | 2 | {2,3}
3 | 2 | local | 45 | 3 | {1,2,3}
(7 rows)
【讨论】：