你能做到的

SELECT MIN(position), id_category
FROM ps_category_product o
WHERE NOT EXISTS (
    SELECT 1
    FROM ps_category_product i 
    WHERE i.position    = o.position + 1
      AND i.id_category = o.id_category
)
GROUP BY id_category;

假设一个可能的位置可以是表中最小位置和最大位置之间的任何位置，并且数据库中的某个位置有一个具有顺序id号的表（在下面的示例中，我有日期），那么您可以创建一个类别可以拥有的所有位置的列表，然后排除它们确实拥有的位置。例如

drop table if exists o;
create table o
(id_category int, id_product int, position int);
insert into o values
(100 ,10 ,0),
(100 ,11 ,1),
(100 ,12 ,2),
(100 ,12 ,3),
(100 ,13 ,4),
(100 ,14 ,5),
(100 ,14 ,6),
(100 ,15 ,7),
(100 ,16 ,8),
(102 ,16 ,8),
(102 ,17 ,9);

select a.id_category, min(possible)
from
(
select o.*,p.possible
from
(
select 0 as possible
union
select id from dates d
where d.id <= (select max(position) from o) + 1
) p
cross join (select distinct id_category from o) o
) a
left join
(select * from o) b on b.id_category = a.id_category and b.position = a.possible
where b.position is null
group by a.id_category;

结果

+-------------+---------------+
| id_category | min(possible) |
+-------------+---------------+
|         100 |             9 |
|         102 |             0 |
+-------------+---------------+
2 rows in set (0.01 sec)