如何计算某个日期的平均结果？

Question

你能帮帮我吗？我有两张桌子。
进入第一个表（活动）有：user_id、sessions 和 login_time。
在第二个（付款）中只有一列 - user_id。

这是我的查询：

SELECT activity.login_time, activity.user_id, avg(activity.sessions) as 
user_sessions
FROM activity
inner JOIN payments ON payments.user_id = activity.user_id
WHERE activity.login_time ='2018-04-05' group by activity.user_id;

使用这个查询，我得到这样的表：

+------------+---------+---------------
| login_time | user_id | user_sessions
+------------+---------+---------------
| 2018-04-05 |     107 |       12.0000
| 2018-04-05 |     110 |        1.0000
| 2018-04-05 |     112 |        5.0000
| 2018-04-05 |     115 |        5.0000
| 2018-04-05 |     117 |        7.0000
| 2018-04-05 |     120 |        1.0000
| 2018-04-05 |     123 |        1.0000
...

我应该如何进行查询以获得平均值：

+------------+------------
| login_time | avg_user_sessions 
+------------+---------
| 2018-04-05 | 4,57

注意：困难在于 user_id 有重复

表格

user_id login_time  sessions
107 2018-04-05  12
110 2018-04-05  1
112 2018-04-05  5
115 2018-04-05  5
117 2018-04-05  7
120 2018-04-05  1
123 2018-04-05  1



user_id
107
107
107
110
112
115
115
117
120
123

Answer 1

如果payments 表中有很多user_id 重复项，您可以尝试在user_id 中的payments 表中使用DISTINCT。

但在你的情况下，你只能直接选择activity，不需要join和payments，因为你没有从中得到任何列。

CREATE TABLE activity(
   login_time date,
   user_id int,
   sessions float
);

CREATE TABLE payments (
   user_id INT
);



INSERT INTO payments VALUES (107);
INSERT INTO payments VALUES (107);
INSERT INTO payments VALUES (110);
INSERT INTO payments VALUES (112);
INSERT INTO payments VALUES (115);
INSERT INTO payments VALUES (115);
INSERT INTO payments VALUES (117);
INSERT INTO payments VALUES (120);
INSERT INTO payments VALUES (123);





INSERT INTO activity VALUES ('2018-04-05',107,12);
INSERT INTO activity VALUES ('2018-04-05',110,1);
INSERT INTO activity VALUES ('2018-04-05',112,5);
INSERT INTO activity VALUES ('2018-04-05',115,5);
INSERT INTO activity VALUES ('2018-04-05',117,7);
INSERT INTO activity VALUES ('2018-04-05',120,1);
INSERT INTO activity VALUES ('2018-04-05',123,1);

查询 1：

SELECT a.login_time, avg(a.sessions) as 
user_sessions
FROM activity a
inner JOIN (SELECT DISTINCT user_id FROM payments) p ON p.user_id = a.user_id
WHERE a.login_time ='2018-04-05' 
group by a.login_time

Results：

| login_time |     user_sessions |
|------------|-------------------|
| 2018-04-05 | 4.571428571428571 |