Question

我确实陷入了困境。

例如我的表是

genre_id  series_id
1         1
1         2
2         5
4         1
2         4
3         3

我希望得到genre_id，其中包含series_id 1 。

因此，我的查询仅返回genre_id 4 而非genre_id 1

但是这个查询

SELECT genre_id FROM test
WHERE genre_id IN (1)
GROUP BY series_id
HAVING COUNT(*) = 1

返回genre_id 1 和 2

Answer 1

/*
drop table t;
create table t (genre_id int, series_id int);
insert into t values
(1,         1),
(1,         2),
(2,         5),
(4,         1),
(2,         4),
(3,         3);
*/

select  s.genre_id
from
(
select   genre_id , max(series_id) maxsid
from        t
group       by genre_id
) s
where   s.maxsid = 1;

结果

+----------+
| genre_id |
+----------+
|        4 |
+----------+

Answer 2

SELECT x.genre_id
  FROM my_table x 
  LEFT 
  JOIN my_table y 
    ON y.genre_id = x.genre_id 
   AND y.series_id NOT IN(1) 
 WHERE y.genre_id IS NULL 
 GROUP 
    BY x.genre_id HAVING COUNT(*) = 1;

Answer 3

好像你使用了错误的查询：

SELECT series_id FROM test WHERE genre_id IN (1) GROUP BY series_id HAVING COUNT(*) = 1

使用

SELECT genre_id FROM test WHERE  series_id IN (1) GROUP BY genre_id HAVING COUNT(*) = 1

Mysql Query正好加工id（s）数组

3 个答案: