集團和過濾mysql的結果

Question

我有以下內容的表：

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

我要計算每個國家的用戶，IP地址，而不是國家的用戶應該算如「未知」 ;

我已成功地寫出下面的SQL查詢：

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

，我得到了以下結果：

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

我怎麼能算所有的IP地址爲「未知」？
我的目標是讓表是這樣的：

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

Answer 1

如果有被視爲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語句與if語句：

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 '%.%.%.%'); 

設置一個變量「未知」所有IP地址