2

The Table I have:

+--------+--------------------+
|item_id|        value        |
+--------+--------------------+
|  1     | 1                  |
|  2     | 4                  |
|  3     | 2                  |
|  4     | 6                  |
+--------+--------------------+

What the SQL Query Result should be: A random combination of items that sum up to 10 with a variable number of different items (in this case 2.)

+--------------+--------------------+-------------+
|item_id       |        amount      |      sum    |
+--------------+--------------------+-------------+
|2             |2                   |8            |
|3             |1                   |2            |
+--------------+--------------------+-------------+

The results show

You get 2 times item 2 (which has the value of 4 so it adds to a sum of 8).

And you get one time item 3 (which has the value of 2)

This combination sums up to 10.

Is that even possible, when it should not be the same combination always and picked random if there are other possibilitys?

4

2 回答 2

1

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;

这会将值放在列上而不是单独的行中。

于 2019-06-30T17:20:03.240 回答
0

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 [较短] 解决方案，自 10.2.2 起可在 MariaDB 中使用（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。

于 2019-06-30T21:58:07.480 回答