by子句

Question

计数不同的值集的同一列在单个组中我有一个表（SQLite的DB）这样，

CREATE TABLE parser (ip text, user text, code text);

现在我需要算code有多少，值为1, 2, or 3，以及有多少不是，由ip组成的字段。

但是，据我所知，我不能完全做到这一点，但有两个SQL短语。

e.g

select count(*) as cnt, ip 
from parser 
where code in (1, 2, 3) 
group by ip 
order by cnt DESC 
limit 10 

并有not in查询。

那么，我可以将两个查询合并成一个单一的？

Answer 1

这会告诉您每ip两项罪名，一个是地方code有值1，2或3，另算为所有的休息（一切，但1，2，3，包括NULL。）

行

SELECT ip, 
     COUNT(CASE WHEN code IN (1, 2, 3) THEN 1 ELSE NULL END) AS cnt_in, 
     COUNT(CASE WHEN code IN (1, 2, 3) THEN NULL ELSE 1 END) AS cnt_rest 
FROM parser 
GROUP BY ip 
ORDER BY cnt_in DESC ; 

这将你给你整数值的休息和第三对具有NULL在code行3个计数，一为1,2,3，另：

SELECT ip, 
     COUNT(CASE WHEN code IN (1, 2, 3) THEN 1 END) AS cnt_in, 
     COUNT(CASE WHEN code NOT IN (1, 2, 3) THEN 1 END) AS cnt_not_in, 
     COUNT(CASE WHEN code IS NULL THEN 1 END) AS cnt_null 
FROM parser 
GROUP BY ip 
ORDER BY cnt_in DESC ; 

---

如果你想限制的第一个结果（如您的代码）前10行，第二个结果到其他前10行，您可以使用两个子查询和UNION：

(SELECT ip, 
     COUNT(*) AS cnt, 
     'in' AS type 
    FROM parser 
    WHERE code IN (1, 2, 3) 
    GROUP BY ip 
    ORDER BY cnt DESC 
    LIMIT 10 
) 
UNION ALL 
(SELECT ip, 
     COUNT(*) AS cnt, 
     'not in' AS type 
    FROM parser 
    WHERE code NOT IN (1, 2, 3) 
    GROUP BY ip 
    ORDER BY cnt DESC 
    LIMIT 10 
) ; 

测试在SQL-Fiddle