【问题标题】:Find the number of unique users who have visited at least two different countries per site找出每个站点访问过至少两个不同国家的唯一用户数
【发布时间】:2019-05-15 14:42:13
【问题描述】:
找出每个网站至少访问过两个不同国家/地区的唯一用户数。
给定时间戳、用户、国家、站点
我认为查询应该是这样的,但它似乎不正确,因为它对每个站点的唯一用户数给出了非常相似的答案。
SELECT site_id, COUNT (DISTINCT user_id)
FROM SWE
GROUP BY site_id
HAVING COUNT(country_id) >=2
ORDER BY site_id ASC;
【问题讨论】:
标签:
mysql
sql
count
distinct
【解决方案1】:
两级聚合是编写查询最自然的方式:
select site_id, count(*)
from (select user_id, site_id, count(*)
from swe
group by user_id, site_id
having min(country) <> max(country) -- or count(distinct country) >= 2
) us
group by site_id;
【解决方案2】:
试试这个-
SELECT A.user_id,B.site_id,COUNT(DISTINCT B.country_id) [Country Visited]
FROM
(
SELECT user_id
FROM SWE
GROUP BY user_id
HAVING COUNT(site_id) = COUNT(DISTINCT site_id)
)A
INNER JOIN SWE B ON A.user_id = B.user_id
GROUP BY A.user_id,B.site_id
HAVING COUNT(DISTINCT B.country_id) >= 2