How to calculate average result for certain date?

How to calculate average result for certain date?



Could you help me? I have two tables.

Into the first table (activity) there are: user_id, sessions and login_time.

Into the second (payments) there's only one column - user_id.



Here's me query:


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;



Using this query, I get such table:


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



How should I make a query to get average:


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



Note: difficulty is in that user_id has duplicates



Tables


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





Why can't you simply group by the date? Pls add sample data and explanation of how the query should work.
– Shadow
Sep 2 at 20:40





@Shadow, could U help me where should I fix it in my query?
– Елисей Горьков
Sep 2 at 20:41




1 Answer
1



If there are many user_id duplicates in payments table, you can try to use DISTINCT in your user_id from payments table.


user_id


payments


DISTINCT


user_id


payments



but in your case, You can only select activity directly, don't need to join with payments, because you didn't get any column from it.


activity


join


payments


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);



Query 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 |





thanks, I tried but this query also displays big list. But I need 1 average value
– Елисей Горьков
Sep 2 at 20:46





Ok could you provide some sample data from your table and show us your expect result from your sample data
– D-Shih
Sep 2 at 20:47





the result must be (according to my example): (12 + 1 + 5 + 5 + 7 + 1 + 1)/7= 4,57
– Елисей Горьков
Sep 2 at 20:50





@ЕлисейГорьков I edit my answer you can try it.
– D-Shih
Sep 2 at 20:56





ok. But this table is formed by my query, because data for this table are in two different tables. That is difficulty
– Елисей Горьков
Sep 2 at 21:00



Thanks for contributing an answer to Stack Overflow!



But avoid



To learn more, see our tips on writing great answers.



Some of your past answers have not been well-received, and you're in danger of being blocked from answering.



Please pay close attention to the following guidance:



But avoid



To learn more, see our tips on writing great answers.



Required, but never shown



Required, but never shown




By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Popular posts from this blog

𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

Crossroads (UK TV series)

ữḛḳṊẴ ẋ,Ẩṙ,ỹḛẪẠứụỿṞṦ,Ṉẍừ,ứ Ị,Ḵ,ṏ ṇỪḎḰṰọửḊ ṾḨḮữẑỶṑỗḮṣṉẃ Ữẩụ,ṓ,ḹẕḪḫỞṿḭ ỒṱṨẁṋṜ ḅẈ ṉ ứṀḱṑỒḵ,ḏ,ḊḖỹẊ Ẻḷổ,ṥ ẔḲẪụḣể Ṱ ḭỏựẶ Ồ Ṩ,ẂḿṡḾồ ỗṗṡịṞẤḵṽẃ ṸḒẄẘ,ủẞẵṦṟầṓế