集团和过滤mysql的结果

Question

我有以下内容的表：

 
| Country  | Username  | 
+---------------+----------------+ 
| US   | John   | 
| IT   | Pedro   | 
| US   | Bob   | 
| FR   | Jill   | 
| 192.168.1.2 | Roupraht  | 
| 192.168.1.20 | Antonio  | 
+---------------+----------------+ 

我要计算每个国家的用户，IP地址，而不是国家的用户应该算如“未知” ;

我已成功地写出下面的SQL查询：

select country, count(*) as total from users group by country; 

，我得到了以下结果：

 
+-----------------+-------+ 
| country   | total | 
+-----------------+-------+ 
| 192.168.1.2  |  1 | 
| 192.168.1.20 |  1 | 
| US    |  2 | 
| IT    |  1 | 
| FR    |  1 | 
+-----------------+-------+ 

我怎么能算所有的IP地址为“未知”？
我的目标是让表是这样的：

 
+-----------------+-------+ 
| country   | total | 
+-----------------+-------+ 
| Unknown   |  2 | 
| US    |  2 | 
| IT    |  1 | 
| FR    |  1 | 
+-----------------+-------+ 

Answer 1

如果有被视为Unknown IP地址，你可以做下面的技巧

select 
country,count(*) as tot 
from users 
where inet_aton(country) is null 
group by country 
union all 
select 
'Unknown',count(*) as tot 
from users 
where inet_aton(country) is not null; 

https://dev.mysql.com/doc/refman/5.0/en/miscellaneous-functions.html#function_inet-aton

Answer 2

，你可以使用mysql LI语句与if语句：

select if(country LIKE '%.%.%.%', 'unknown', country), count(*) as total from users group by country; 

Answer 3

您可以使用此

set @unknown = (select country from table where country LIKE '%.%.%.%'); 

设置一个变量“未知”所有IP地址