我想为每天发布超过一件事的用户获取所有USER_ID,
我原来尝试过
SELECT USER_ID, count(DISTINCT cast(POSTING_DATE as DATE))
AS NUM_DAYS_OF_DUPLICATES FROM POSTING_TABLE
WHERE USER_ID IN
(SELECT USER_ID FROM POSTING_TABLE
GROUP BY CAST(POSTING_DATE AS DATE) HAVING count(*) >= 2)
GROUP BY USER_ID ORDER BY NUM_DAYS_OF_DUPLICATES DESC;
然后这适用于特定的USER_ID
SELECT USER_ID FROM POSTING_TABLE WHERE USER_ID = 30
GROUP BY cast(POSTING_DATE AS DATE)
HAVING count(cast(POSTING_DATE AS DATE)) > 1
上面给了我正确的结果,但是当我在整个表上运行查询而不指定USER_ID时却没有。
例如。,
表结构USER_ID,POSTING_DATE ...
USER_ID POSTING_DATE
1 10-10-13
1 10-10-13
1 10-12-13
1 10-12-13
2 10-10-13
2 10-10-13
3 10-10-13
4 10-12-13
结果会给我什么
USER_ID NUM_DAYS_WITH_MORE_THAN_ONE_POSTING
1 2
2 1
3 0
4 0
另外,如果我们可以省略0
最佳答案
这是解决方案
select x.user_id, count(x.num_days)
from
(
select USER_ID, COUNT(USER_ID) AS NUM_DAYS
from data1
group by user_id, posting_date
having count(user_id) > 1
) x
group by 1
Working SQL Fiddle
(为简单起见,我使用了varchar作为日期,但它也可以与
date
一起使用。您可以检查自己的数据库)关于mysql - MySQL查询选择每天有多个条目的所有用户,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/17555576/