只有同時使用DISTINCT和GROUP BY

Question

當計一審

說我有如下表，

timedate id 
    2015-01-01 1 
    2015-01-01 2 
    2015-01-01 3 
    2015-01-01 4 
    2015-01-02 1 
    2015-01-02 2 
    2015-01-02 5 
    2015-01-02 6 
    2015-01-03 2 
    2015-01-03 3 

這個查詢

SELECT 
     COUNT(DISTINCT `id`) as total, 
     timedate 
FROM Table1 
GROUP BY timedate 

產生以下結果

total timedate 
    4  2015-01-01 
    4  2015-01-02 
    2  2015-01-03 

因爲它計算每個組內不同的id。

我該如何獲得每個不同ID的計數，而不在隨後的組中進行計數？例如像這樣的結果：

total timedate 
    4  2015-01-01 
    2  2015-01-02 
    0  2015-01-03 

Answer 1

你最好在這裏使用兩個子查詢。一拿到id和它的min(timedate)，另一個是得到不同timedates：

SELECT 
     COUNT(DISTINCT `id`) as total, 
     t1.timedate 
FROM 
    (SELECT DISTINCT TimeDate FROM Table1) as t1 
    LEFT OUTER JOIN (SELECT min(Timedate) as firsttimedate, id FROM Table1 GROUP BY id) as t2 ON 
     t1.timedate = t2.firsttimedate 
GROUP BY t1.timedate