SQL返回合并的结果集

Question

我有以下SQL:

SELECT `table1`.`value`, `table2`.* 
FROM `table2` 
INNER JOIN `table1` ON `table2`.`product_id` = `table1`.`entity_id`
WHERE `table2`.`created_at` > '2012-04-23' and 
(`table1`.`value` = 264 OR `table1`.`value` = 260)
order by order_id

返回一个这样的结果集(这只是返回结果的一部分):

value  order_id   ...
260    1234
260    1235
260    1236
264    1236
260    1237
260    1238
260    1239
264    1239
264    1240
260    1241

我想要的是一个查询,它将获取这些结果,并且只返回order_id包含值260和264的订单.基于此示例,我要查找的最终结果是

260   1236
264   1236
260   1239
264   1239

我的想法是这可以用一个子集完成,但我不完全确定完成它的细节.

Answer 1

这可以通过关系划分来完成:

select r.order_id from (
  select 
    dividend.*  
  from your_table_or_query as dividend  -- assumes no duplicates in `dividend`; use `distinct` if there are any
  inner join divisor
  on dividend.value = divisor.value
) as r
group by r.order_id
having count(*) = (select count(*) from divisor);

结果:

+----------+
| order_id |
+----------+
|     1236 |
|     1239 |
+----------+
2 rows in set (0.00 sec)

您的查询所在的位置your_table_or_query和

select 260 as value from dual union select 264 as value from dual

是divisor.

这将返回订单ID 1236和1239; 然后,可以将它们join编辑为原始查询,以获取具有这些订单ID的所有行,如果这是您想要的.

完整查询以及insert语句:

create table divisor (value int);
insert into divisor values (260), (264);

create table your_table_or_query (value int, order_id int);
insert into your_table_or_query values (260, 1234), (260, 1235), (260, 1236), (264, 1236), (260, 1237), (260, 1238), (260, 1239), (264, 1239), (264, 1240), (260, 1241);


select y.* from (
  select r.order_id from (
    select
      dividend.*
    from your_table_or_query as dividend
    inner join divisor
      on dividend.value = divisor.value
  ) as r 
  group by r.order_id
  having count(*) = (select count(*) from divisor)
) as quotient
inner join your_table_or_query y
  on quotient.order_id = y.order_id;

结果:

+-------+----------+
| value | order_id |
+-------+----------+
|   260 |     1236 |
|   264 |     1236 |
|   260 |     1239 |
|   264 |     1239 |
+-------+----------+
4 rows in set (0.00 sec)