by子句

Question

計數不同的值集的同一列在單個組中我有一個表（SQLite的DB）這樣，

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

現在我需要算code有多少，值爲1, 2, or 3，以及有多少不是，由ip組成的字段。

但是，據我所知，我不能完全做到這一點，但有兩個SQL短語。

e.g

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

並有not in查詢。

那麼，我可以將兩個查詢合併成一個單一的？

Answer 1

這會告訴您每ip兩項罪名，一個是地方code有值1，2或3，另算爲所有的休息（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 ; 

這將你給你整數值的休息和第三對具有NULL在code行3個計數，一爲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個結果（如您的代碼）前10行，第二個結果到其他前10行，您可以使用兩個子查詢和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 
) ; 

測試在SQL-Fiddle