PostgreSQL數據庫

Question

我想，以使該選擇必須使用PostgreSQL數據庫

Answer 1

像這樣的東西應該做的伎倆相同的用戶名和創建日期的同一時刻用戶的查詢。這將返回任何用戶/小時對與計（未經測試）沿：

select users.username, datepart('hour', users.created_at), count(*) from users 
    inner join users u2 
    on users.username = u2.username 
    and datepart('hour', users.created_at) = datepart('hour', u2.created_at) 
group by users.username, datepart('hour', users.created_at) having count(*) > 1 

Answer 2

select u.* 
from users u 
join (
    select username, date_trunc('hour', creation_timestamp) 
    from users 
    group by 1, 2 
    having count(*) > 1 
) as x on u.username = x.username 
order by u.username; 

應該很好地工作。