我有一个表(user_admin_password)。我正在使用此查询来获取记录。
SELECT user_key,admin_status,count(*)
FROM user_admin_status
GROUP BY
user_key,admin_status having count(*) > 1
ORDER BY user_key,admin_status;
结果是:
+----------+--------------+----------+
| user_key | admin_status | count(*) |
+----------+--------------+----------+
| 1 | NON-DBA | 5 |
| 3 | DBA | 328 |
| 5 | NON-DBA | 8 |
| 6 | NON-DBA | 25 |
| 7 | NON-DBA | 4 |
| 9 | DBA | 232 |
| 10 | NON-DBA | 4 |
| 11 | DBA | 4 |
| 13 | NON-DBA | 8 |
| 15 | NON-DBA | 2 |
| 16 | DBA | 326 |
| 16 | NON-DBA | 2 |
| 17 | NON-DBA | 10 |
| 18 | NON-DBA | 5 |
| 19 | NON-DBA | 12 |
| 20 | NON-DBA | 2 |
| 21 | NON-DBA | 2 |
...
...
现在,我想要所有带有重复记录的user_keys ...例如16
...任何建议???
答案 0 :(得分:3)
将其包装在另一个选择中:
select user_key, count(*)
from
(
select user_key,admin_status,count(*)
from user_admin_status
group by user_key,admin_status
having count(*) > 1
) x
group by user_key
having count(*) > 1