如何计算特定日期的平均结果？

Question

你可以帮帮我吗？ 我有两张桌子。
进入第一个表（活动），有：user_id，会话和login_time。
在第二（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;

使用此查询，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
...

我应该如何查询以获得平均值：

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

注意：问题在于user_id有重复项

表

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

如果有很多user_id在重复payments表，你可以尝试使用DISTINCT在你user_id从payments表。

但在您的情况下，您只能直接选择activity ，而无需join payments ，因为您没有从中获取任何列。

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

结果 ：

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