如何获得总计为特定值的行的随机组合?

How to get a random combination of rows that sum up to a specific value?

我的Table:

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

SQL 查询结果应该是什么: 项目的随机组合,总和为 10,不同项目的数量可变(在本例中为 2。)

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

结果显示

您得到 2 次项目 2(它的值为 4,所以它的总和为 8)。

并且您获得了一次项目 3(其值为 2)

这个组合总和为 10。

这甚至可能吗,如果有其他可能性,它不应该总是相同的组合并且随机选择吗?

您可以使用 self-join:

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;

这会将值放在列中而不是单独的行中。

假设你想要一个单一的随机组合,你可以这样做:

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

每次你运行这个查询它都会选择一个随机的[不同]解决方案。

创建您提到的 table 和数据(我曾经测试过)的脚本是:

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,这是一个使用递归 CTE(通用 Table 表达式)的等效 [较短] 解决方案,自 MariaDB 以来可用10.2.2(见Recursive Common Table Expressions):

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

如果您需要使用更大的数字,此解决方案的扩展性会更好。