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TA贡献1826条经验 获得超6个赞
您可以使用自联接获得所有此类组合:
select t1.item_id, t2.item_id
from t t1 join
t t2
on t1.value + t2.value = 10;
这会将值放在列而不是单独的行中。
TA贡献1853条经验 获得超6个赞
假设您想要一个随机组合,您可以执行以下操作:
select
*
from (
select
a.item_id as item1,
x.n as amount1,
a.value * x.n as sum1,
b.item_id as item2,
y.n as amount2,
b.value * y.n as sum2,
rand() as r
from my_table a
join my_table b on b.item_id <> a.item_id
cross join (
select 1 as n union select 2 union select 3 union select 4
union select 5 union select 6 union select 7 union select 8
union select 9 union select 10) x
cross join (
select 1 as n union select 2 union select 3 union select 4
union select 5 union select 6 union select 7 union select 8
union select 9 union select 10) y
where a.value * x.n + b.value * y.n = 10
) z
order by r -- sorted randomly
limit 1 -- to get only one combination; remove to get them all
每次运行此查询时,它都会选择一个随机的 [不同] 解决方案。
创建您提到的表和数据的脚本(我曾经测试过)是:
create table my_table (
item_id int,
value int
);
insert into my_table (item_id, value) values (1, 1);
insert into my_table (item_id, value) values (2, 4);
insert into my_table (item_id, value) values (3, 2);
insert into my_table (item_id, value) values (4, 6);
2019 年 7 月 1 日编辑:根据要求,这是使用递归 CTE(公用表表达式)的等效 [更短] 解决方案,自 10.2.2 起在 MariaDB 中可用(请参阅递归公用表表达式):
with recursive
val as (select 1 as n union all select n + 1 from val where n < 10)
select
*
from (
select
a.item_id as item1,
x.n as amount1,
a.value * x.n as sum1,
b.item_id as item2,
y.n as amount2,
b.value * y.n as sum2,
rand() as r
from my_table a
join my_table b on b.item_id <> a.item_id
cross join val x
cross join val y
where a.value * x.n + b.value * y.n = 10
) z
order by r -- sorted randomly
limit 1 -- to get only one combination; remove to get all 22 answers
如果您需要使用更高的数字,此解决方案的扩展性会更好。
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