Question

我在围绕这个选择声明缠绕我的大脑时遇到了麻烦。数据来自3个表格（为了便于阅读，我删除了所有不必要的数据）：

mysql> describe vulnerability;
+---------------+------------------+------+-----+---------+----------------+
| Field         | Type             | Null | Key | Default | Extra          |
+---------------+------------------+------+-----+---------+----------------+
| vuln_id       | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| severity      | int(10) unsigned | NO   |     | NULL    |                |
| host_id       | int(10) unsigned | NO   | MUL | NULL    |                |
+---------------+------------------+------+-----+---------+----------------+

mysql> describe cve;
+---------+------------------+------+-----+---------+----------------+
| Field   | Type             | Null | Key | Default | Extra          |
+---------+------------------+------+-----+---------+----------------+
| cve_id  | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| cve     | varchar(15)      | NO   |     | NULL    |                |
| vuln_id | int(10) unsigned | NO   | MUL | NULL    |                |
| year    | int(4) unsigned  | YES  |     | NULL    |                |
+---------+------------------+------+-----+---------+----------------+

mysql> describe host;
+--------------+------------------+------+-----+---------+----------------+
| Field        | Type             | Null | Key | Default | Extra          |
+--------------+------------------+------+-----+---------+----------------+
| host_id      | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| ip_addr      | int(10) unsigned | NO   |     | NULL    |                |
+--------------+------------------+------+-----+---------+----------------+

我想输出漏洞数量少于2009年且严重性为3的主机数量。年份包含在CVE中，与漏洞FK的漏洞相关联。该漏洞具有严重性，并与host_id FK绑定到Host。以下是我到目前为止的情况：

mysql> select count(distinct ip_addr) from host H 
  inner join vulnerability V on H.host_id = V.host_id 
  inner join CVE C on C.vuln_id = V.vuln_id 
  where V.severity = 3 and C.year < 2009;
+-------------------------+
| count(distinct ip_addr) |
+-------------------------+
|                    5071 |
+-------------------------+

这告诉我有超过2009年的漏洞的主机总数，这是一个良好的开端。但是，我想更进一步，只包括那些有50个或更多漏洞的主机。我不知道该怎么做。 Host表中的每个主机条目都有多个相应的漏洞条目。我想我需要在where子句中添加一些内容，但是我被卡住了。

提前致谢。如果需要更多信息，请告诉我。

Answer 1

尝试使用GROUP BY和HAVING：

SELECT ip_addr
FROM host AS H
INNER JOIN vulnerability AS V
    ON H.host_id = V.host_id
INNER JOIN CVE AS C
    ON C.vuln_id = V.vuln_id
WHERE V.severity = 3 AND C.year < 2009
GROUP BY ip_addr
HAVING COUNT(DISTINCT vuln_id) >= 50

要获取计数，请将上述查询包装在另一个查询中：

SELECT COUNT(*) FROM
(
     SELECT ip_addr
     FROM host AS H
     -- etc... same query as above
) T1

Mysql：从多个连接表中选择特定数据

1 个答案: