如何GROUP BY與例外？

Question

這裏是我的簡單求代碼：

SELECT 
    a.user_id as User_ID, 
    min(b.a_day) as Date_from, 
    max(b.a_day) as Date_to, 
    c.code as ID 
FROM a, b, c 
WHERE 
    a_day > (day, -15, getdate()) 
GROUP BY 
    a.user_id, 
    c.code 

查詢提供了以下的輸出：

User ID date_from date_to  id  
1234567 2016-06-13 2016-06-13 B 
1234567 2016-06-17 2016-06-17 A 
12345672016-06-18 2016-06-18 A 
1234567 2016-06-19 2016-06-19 A 
1234567 2016-06-20 2016-06-20 A 
1234567 2016-06-21 2016-06-21 C 
1234567 2016-06-22 2016-06-22 C 
1234567 2016-06-23 2016-06-23 D 

我需要的是這樣的：

User ID date_from date_to  id 
1234567 2016-06-13 2016-06-13 B 
1234567 2016-06-17 2016-06-20 A 
1234567 2016-06-21 2016-06-22 C 
1234567 2016-06-23 2016-06-23 D 

當我使用min（）和max（ ）函數與分組相同，它彙總了所有記錄的罰款，但我必須每天只彙總具有相同ID的日期。

任何想法？

在此先感謝。

Answer 1

您可以通過GROUP BY子句中使用CASE EXPRESSION有條件的分組做到這一點：

SELECT 
    a.user_id as User_ID, 
    min(b.a_day) as Date_from, 
    max(b.a_day) as Date_to, 
    c.code as ID 
FROM a, b, c 
WHERE 
    a_day > (day, -15, getdate()) 
GROUP BY 
    a.user_id, 
    c.code, 
    CASE WHEN c.code in ('B','D') THEN b.a_day ELSE 1 END 

將生成以此爲GROUP BY條款：

c.code = 'B' -> a.user_id,c.code,b.a_day 
c.code <> 'B' -> a.user_id,c.code,1