Question

我有一个查询问题，想知道这是否可以查询。在我提出问题的数据库看起来像这样：

-------------------------------------------
id  |  name       |        IP      |   nr
-------------------------------------------
 1  | Fleet-1234  | 192.168.52.254 | 1234
 5  | Boat-4567   | 192.168.52.256 | NULL
 9  | Boat-4565   | 192.168.52.250 | NULL
10  | Boat-4547   | 192.168.52.252 | NULL
13  | Fleet-6663  | 192.168.44.256 | 6663
18  | Boat-2345   | 192.168.44.145 | NULL
56  | Boat-2345   | 192.168.44.14  | NULL
60  | Boat-1111   | 192.168.44.15  | NULL

我希望的结果是：

-------------------------------------------
id  |  name       |        IP      |   nr
-------------------------------------------
 1  | Fleet-1234  | 192.168.52.254 | 1234
 5  | Boat-4567   | 192.168.52.256 | 1234
 9  | Boat-4565   | 192.168.52.250 | 1234
10  | Boat-4547   | 192.168.52.252 | 1234
13  | Fleet-6663  | 192.168.44.256 | 6663
18  | Boat-2345   | 192.168.44.145 | 6663
56  | Boat-2345   | 192.168.44.14  | 6663
60  | Boat-1111   | 192.168.44.15  | 6663

我认为我需要做的是： - 比较IP的前3个八位字节（相同的＃34;车队＆＃34;） -iff比较被发现使用＆＃34; nr＆＃34;来自舰队-XXXX的colum在同一＆＃34;组＆＃34;

我真的不知道这是否可行以及如何做到这一点。是否有人可以帮助我查询此查询？

Answer 1

请参阅sqlfiddle：

SELECT 
    t0.id
    , t0.name
    , t0.IP
    , t1.nr
FROM your_table t0
INNER JOIN (SELECT LEFT(IP, LENGTH(IP) - LOCATE('.', REVERSE(IP))) AS shortened_IP, MAX(nr) AS nr FROM your_table GROUP BY 1) t1
ON LEFT(IP, LENGTH(IP) - LOCATE('.', REVERSE(IP))) = t1.shortened_IP;

Answer 2

使用substr和group by

select substr(IP, LOCATE('.', REVERSE(IP))+1, 3) as ip_base, count(*)
from your_table 
group by ip_base;

进行查询以按IP分组

2 个答案: