如何通過一年

Question

我想通過一年來算兩場

我嘗試

SELECT YEAR(admit_date),COUNT(sub_case),COUNT(SK) 
FROM admit 
WHERE sub_case = "scase001" 
    AND SK = "rsk002" 
GROUP BY YEAR(admit_date),sub_case,SK 

但要算兩場它不是真正的

表

總

真

2014 1 0 
2015 1 1 

Answer 1

刪除sub_case和SK列，您可以使用條件的聚集，而不是像這樣：

SELECT 
    YEAR(admit_date) AS "admit_year", 
    SUM(sub_case = 'scase001') AS "sub_case_count", 
    SUM(SK = 'rsk002') AS "sk_count" 
FROM admit 
GROUP BY YEAR(admit_date) 

表達式對於所有匹配的行，將評估爲1，並且總結匹配與計數它們相同。以這種方式使用（或濫用）布爾表達式可能無法在其他SQL數據庫中工作，但它在MySQL中（通常會使用case when ... then .. end表達式）。

Answer 2

您需要從GROUP BY

SELECT YEAR(admit_date),COUNT(sub_case),SUM(CASE WHEN SK="rsk002" THEN 1 ELSE 0 END) 
FROM admit 
WHERE sub_case = "scase001" 
GROUP BY YEAR(admit_date)