select 日名称和最大平均值如何与 sql 的数据库

Question

I have the following columns in my database: - day - browser - platforms - visitors_number

I have to choose the day of the week for which the average number of visitors was maximum and display this maximum average. I searched the forum, found a similar thread, but the solution does not work, I get an error. I would add that I am sitting on it for the second day and I lack ideas. I will add that it works on phpMyAdmin.

This is my code:

    select DAYNAME(DAY), avgVIS
    from 
    (
    select DAYNAME(DAY), avg(VISITORS_NUMBER) as avgVIS
    from dane_2 
    group by DAYNAME(DAY)
    ) 
    where avgVIS = (select max(avgVIS) 
              from ( select DAYNAME(DAY), avg(VISITORS_NUMBER) as avgVIS
              from dane_2 
              group by DAYNAME(DAY)))

this is output's error

#1064 - Something is wrong in your syntax obok 'where avgVIS = (select max(avgVIS)
              from ( select DAYNAME(DAY), a'

Can someone helps me? Thank you in advance

Answer 1

You already have the aggregation query that computes the daily average. All that is let to do is sort and limit:

select dayname(d.DAY) day_name, avg(d.VISITORS_NUMBER)  avg_vis
from dane_2 d
group by dayname(d.DAY) 
order by avg_vis desc
limit 1

Note: DAY is a reserved word in MySQL, so not a good choice for a column name (I qualified it with the table alias to avoid errors).