SQL通过分组计数行

Question

我有一张桌子：

A       B
 ---------
1       Date1
1       Date1
1       Date1
1       Date2
2       Date1
3       Date3
4       Date2
4       Date1
1       Date3

SELECT A, count(A) FROM `table` WHERE B BETWEEN 'Date1' AND 'Date2'
GROUP BY A
ORDER BY A

当我进行正常查询时：

A       B
----------
1        1
1        1
1        1
2        1
3        1
4        1
4        1

需要算上几天？对此有任何简单的疑问吗？请帮忙

需要以下数据：

A       B
----------
1        3
2        1
3        1
4        2

Answer 1

似乎你需要数（不同的B）

SELECT A, count(distinct B) 
FROM `table` WHERE B BETWEEN 'Date1' AND 'Date2'
GROUP BY A 
ORDER BY A;

Answer 2

问题解决了：

SELECT T1.id as id, COUNT(COALESCE(T2.totalcountdepartmentuserwise, 0)) AS totalcountdepartmentuserwise 
FROM (
      SELECT DISTINCT id FROM users WHERE company_id = 33
     ) AS T1 
     LEFT JOIN (
        SELECT user_id, COUNT(distinct user_id) AS totalcountdepartmentuserwise FROM userlogs WHERE created_at BETWEEN '2016-01-01' AND '2016-01-25' GROUP BY user_id, DATE(created_at) ORDER BY user_id
     ) AS T2 ON T2.user_id = t1.id GROUP BY id