MySQL查詢幫助

Question

我有以下模式的日誌表：

OperatorId - JobId - Status (Good/Bad/Ugly) 
Alex   6  Good 
Alex   7  Good 
James  6  Bad 

說明：當操作員工作在一個工作，一個條目與狀態一起做。而已。

現在我需要像一個報告：

OperatorId - Good Count - Bad Count - Ugly Count 
Alex   2   0   0 
James  0   1   0 

幫助！

Answer 1

select operatorid, 
sum(if(status="good",1,0)) as good, 
sum(if(status="bad",1,0)) as bad, 
sum(if(status="ugly",1,0)) as ugly 
from table 
group by operatorid 

Answer 2

這被稱爲數據透視表。它通過設定各狀態的值1或0，然後總結起來做：

SELECT 
    T.OperatorId, 
    SUM(T.GoodStat) AS Good, 
    SUM(T.BadStat) AS Bad, 
    SUM(T.UglyStat) AS Ugly 
FROM 
(
SELECT 
    CASE WHEN Status = 'Good' THEN 1 ELSE 0 END AS GoodStat, 
    CASE WHEN Status = 'Bad' THEN 1 ELSE 0 END AS BadStat, 
    CASE WHEN Status = 'Ugly' THEN 1 ELSE 0 END AS UglyStat, 
    OperatorId 
    FROM logTable T 
) 
GROUP BY T.OperatorId 

Answer 3

如果像我一樣，你喜歡，只要有可能，計算與計數時，而不是SUM，這裏是一個替代解決方案，它使用this thread詢問的方法：

SELECT 
    operatorid, 
    COUNT(status = 'good' OR NULL) as good, 
    COUNT(status = 'bad' OR NULL) as bad, 
    COUNT(status = 'ugly' OR NULL) as ugly 
FROM table 
GROUP BY operatorid