Question

我有下表

video_id    ip_address
1           192.168.1.1
1           192.168.1.1
1           192.168.1.2
2           192.168.1.2
2           192.168.1.2
2           192.168.1.2

运行以下查询将返回每个IP地址的每个视频的唯一请求数量。

SELECT video_id, COUNT(*) as num_unique_requests FROM 
    (SELECT video_id, ip_address FROM videos
    GROUP BY video_id, ip_address) t
GROUP BY video_id
ORDER BY num_unique_requests DESC

video_id    num_unique_requests
1           2
2           1

有没有办法可以将最里面的GROUP BY语句与另一个GROUP BY语句嵌套，这样我就可以避免创建最外层的语句GROUP BY video_id?

还有其他方法可以更有效地编写此查询吗？

Answer 1

只需使用 COUNT（DISTINCT FILEDNAME）

SELECT video_id, COUNT( DISTINCT ip_address) as num_unique_requests FROM 
videos
GROUP BY video_id
ORDER BY num_unique_requests DESC

Answer 2

在您的情况下，内部查询中的DISTINCT是apt，代替GROUP BY

SELECT video_id, COUNT(*) as num_unique_requests 
FROM 
    (SELECT DISTINCT video_id, ip_address FROM videos) t
GROUP BY video_id
ORDER BY num_unique_requests DESC

是否可以将GROUP BY语句嵌套在彼此内？

2 个答案: