请帮帮我,我正在尝试解决这个问题很长时间...... 我有表产品和相关产品(顶级产品由其他基础产品组成(。目标:我想得到所有的基础产品。 因此,表如下所示:
product_id related_product_ids
------------------------------------------------
1143 1213
1255 1245
1261 1229,1239,1309,1237,1305,1243,1143
我通过查询得到了这个:
select max(p.id) as product_id, array_to_string(array_agg(p2p.related_product_id), ',') as related_product_ids
from product p
left join product_to_product p2p on p2p.product_id = p.id
where p.id in (select product_id from order_line where wo_id = 262834)
group by p.id, p2p.product_id
我想将related_product_ids
输入到产品表中以获取所有相关产品。 所以,实际上我通过运行从所有必要的值制作数组
select array_agg(p2p.related_product_id) as id
from product p
left join product_to_product p2p on p2p.product_id = p.id
where p.id in (select product_id from order_line where wo_id = 262834)
related_product_ids
---------------------------------------------
{1309,1143,1229,1239,1243,1237,1305,1245,1213}
我尝试了以下方法,但没有成功:
select *
from product
where id = ANY(select array_agg(p2p.related_product_id) as id
from product p
left join product_to_product p2p on p2p.product_id = p.id
where p.id in (select product_id from order_line where wo_id = 262834))
Error: ERROR: operator does not exist: integer = integer[] Hint: No operator matches the given name and argument type(s). You might need to add explicit type casts. Position: 39, SQLState: 42883, ErrorCode: 0
或以下:
select *
from product
where id in (select array_to_string(array_agg(p2p.related_product_id), ',') as id
from product p
left join product_to_product p2p on p2p.product_id = p.id
where p.id in (select product_id from order_line where wo_id = 262834))
Error: ERROR: operator does not exist: integer = integer[] Hint: No operator matches the given name and argument type(s). You might need to add explicit type casts. Position: 36, SQLState: 42883, ErrorCode: 0
和许多其他尝试
所以最后我需要的是
select *
from product
where id in (1309,1143,1229,1239,1243,1237,1305,1245,1213)
(值来自related_product_ids
(
如何将整数数组 (related_product_ids( 转换为值....或者你可以建议不同的更好的方法?
DBFiddle
如果你想将结果用作数组,你可以用ANY
来做到这一点 - 但参数也必须是一个数组。
select *
from product
where id = any(array(select p2p.related_product_id
from product p
left join product_to_product p2p on p2p.product_id = p.id
where p.id in (1, 2, 3)))
但我认为你把事情复杂化了。据我所知,这可以简化为:
select p1.*
from product p1
where exists (select *
from product_to_product p2p
where p2p.related_product_id = p1.id
and p2p.product_id in (1,2,3))
目标:我想得到所有的基础产品。
如果我假设"基本"产品是从未出现在related_product_id
列中的产品,那么not exists
想到:
select p.*
from product p
where not exists (select 1
from product_to_product p2p
where p2p.related_product_id = p.id
);
我不知道为什么你的 =ANY 不起作用,在我看来它应该。因为 select 理论上可以返回多行,所以它将array_agg视为嵌套数组的内部数组。 ANY "取消嵌套"第一层,但仍留下一个 int[] 层供=
使用。
但是,如果您只是摆脱聚合,您的 IN 示例就可以工作:
由于您没有为表提供创建脚本,因此我替换了 pgbench 中的脚本,以便我可以发布经过测试的代码。 该概念应适用于您的表。
select * from pgbench_accounts where aid in (select bid from pgbench_branches);
请注意,当您不聚合时,ANY 也有效:
select * from pgbench_accounts where aid =ANY (select bid from pgbench_branches);
列表、数组和集合是不同的东西。 但在某些情况下,它们可以互换使用。 但是我不知道如何在不尝试的情况下预测哪些。
DBFiddle 示例中的错误是:
在最后一个查询中,只需unnest
数组而不是array_to_string
select * from product where id = ANY(select unnest(array_agg(p2p.related_product_id)) as id from product p
left join product_to_product p2p on p2p.product_id = p.id
where p.id in (1, 2, 3))