Question

我有一张这样的表：

Hour    Operation   IP
10  GET     192.168.1.3
10  POST    192.168.1.3
10  GET     192.168.1.4
10  GET     192.168.1.8
12  POST    192.168.1.7
12  GET     192.168.1.7
12  GET     192.168.1.9

我需要知道每个小时有多少不同的GET操作没有关联的POST操作（没有购买的访问）

答案是：

Hour GETs
10 2
12 1

但我不清楚如何做到这一点。

Answer 1

嗯。嗯。。。我想你想要：

select hour, sum(case when posts > gets then gets - posts else 0 end)
from (select ip, hour,
             sum(case when operation = 'GET' then 1 else 0 end) as gets,
             sum(case when operation = 'POST' then 1 else 0 end) as posts
      from t
      group by ip, hour
     ) ih
group by hour;

这是相当棘手的，因为大概给定的IP可能会在同一时间内有多个获取和放置。

Answer 2

这应该有效：

 SELECT HOUR, COUNT(IP) FROM 
 (
   SELECT HOUR, IP FROM mytable a WHERE Operation = 'GET' and NOT EXISTS
    ( SELECT 1 FROM mytable b WHERE Operation = 'POST' AND a.IP = b.IP AND 
      a.HOUR = b.HOUR
    )
) as a
GROUP BY HOUR

甚至更短：

SELECT HOUR, COUNT(IP) 
FROM mytable a 
WHERE Operation = 'GET' 
AND NOT EXISTS
(
  SELECT 1 FROM mytable b 
  WHERE Operation = 'POST' 
  AND a.IP = b.IP 
  AND a.HOUR = b.HOUR
)
GROUP BY HOUR

获取的SQL分组与任何帖子无关

2 个答案: