如何計算特定日期的平均結果？

Question

你可以幫幫我嗎？ 我有兩張桌子。
進入第一個表（活動），有：user_id，會話和login_time。
在第二（付款）中，只有一列-user_id。

這是我的查詢：

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;

使用此查詢，我得到這樣的表：

+------------+---------+---------------
| 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
...

我應該如何查詢以獲得平均值：

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

注意：問題在於user_id有重復項

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在重復payments表，你可以嘗試使用DISTINCT在你user_id從payments表。

但在您的情況下，您只能直接選擇activity ，而無需join payments ，因為您沒有從中獲取任何列。

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);

查詢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

結果 ：

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