Question

表“A”内容：

      IP       | Port
----------------+-------
10.211.240.300 | 10
10.211.240.300 | 10
10.211.240.300 | 20
10.211.240.300 | 20
10.211.240.300 | 20
10.211.240.300 | 20
10.211.240.300 | 30
10.211.240.300 | 30
10.211.240.300 | 30
10.211.240.200 | 10
10.211.240.200 | 10
10.211.240.200 | 10
10.211.240.200 | 20
10.211.240.200 | 20
10.211.240.200 | 20
10.211.240.200 | 20
10.211.240.200 | 30
10.211.240.100 | 10
10.211.240.100 | 10
10.211.240.100 | 10
10.211.240.100 | 10
10.211.240.100 | 10
10.211.240.100 | 10
10.211.240.100 | 10
10.211.240.100 | 10
10.211.240.100 | 20
10.211.240.100 | 20
10.211.240.100 | 20
10.211.240.100 | 30
10.211.240.100 | 30
10.211.240.100 | 30
10.211.240.100 | 30
10.211.240.100 | 30

我需要为其地址订购IP地址，并且还需要对其端口进行有序计数，因此上述数据的预期结果是：

       IP      | Port    | PortCount  | TotalCount
----------------------------------------------
 10.211.240.100 | 10      |  08        | 16
 10.211.240.100 | 30      |  05        | 16
 10.211.240.100 | 20      |  03        | 16
 10.211.240.300 | 30      |  04        | 09
 10.211.240.300 | 20      |  03        | 09
 10.211.240.300 | 10      |  02        | 09
 10.211.240.200 | 20      |  04        | 08
 10.211.240.200 | 10      |  03        | 08
 10.211.240.200 | 30      |  01        | 08

有人可以帮我解决postgres关系查询吗？

Answer 1

SELECT A.IP, A.PORT, T2.PORT_COUNT, COUNT(*) 
FROM A JOIN (
             SELECT T1.IP, COUNT(T1.PORT) AS PORT_COUNT
             FROM A AS T1
             GROUP BY T1.IP  
            ) AS T2
            ON A.IP =  T2.IP
GROUP BY A.IP, A.PORT

我测试过的在线数据库：http://sqlfiddle.com/#!9/23a78/6

Answer 2

如果您使用的是postgresql，请使用查询： -

SELECT a.IP, a.PORT,total_count.total,COUNT(*)
FROM a
JOIN (
         SELECT b.IP, COUNT(b.PORT) AS total
         FROM a AS b
         GROUP BY b.IP  
        ) AS total_count
        ON a.IP =  total_count.IP
GROUP BY a.IP, a.PORT,total_count.total
ORDER BY a.ip

分组已经分组和聚合的数据 - Postgresql 9.3及以上

2 个答案: