How to sum same rows and find those that are above average in mySQL?

Question

This is a sample of my database table:

visitor_id ---- url ---- duration
1 ------------ home ------- 5
1 ------------ about ------ 8
1 ------------ about ------ 3
1 ------------ contact ---- 2
1 ------------ home ------- 3
1 ------------ services --- 2

What I want to achieve is to show the urls that are more than the average of all urls. But first I need to sum the duration group by url.

For example, the average of all pages = 5+8+3+2+3 + 2 = 23 / 6 = 3.8

So the pages that will be shown is home ( 8 secs) and about (11).

How can I achieve this?

This is what I have so far

select sum(duration), url from pages where visitor_id='1' group by url

Answer 1

You can join in the average or use a having clause. I would write this as:

select p.visitor_id, p.url, avg(p.duration)
from pages p
where visitor_id = 1 
group by p.visitor_id, p.url
having avg(p.duration) > (select avg(p2.duration)
                          from pages p2
                          where p2.visitor_id = p.visitor_id
                         );

Notes:

• I'm guessing that visitor_id is numeric, so the constant is a number instead of a string.
• The subquery in the having clause is correlated to the outer query, so the visitor id is input only once.
• The query aggregates by both visitor id and url, so it will work for any number (or all) visitors.

Answer 2

You could use an having on subquery for avg

  select sum(duration) sum_dur, url 
  from pages 
  where visitor_id='1' 
  group by url 
  having sum_dur > (
  select avg(duration)  
  from  pages 
  where visitor_id='1' 
  )

Answer 3

Try this:

Select url
,visitor_id
,duration
From (
select url
,visitor_id
,sum(duration) as duration
,avg(duration) over (partition by visitor_id) as avg_by_visitor_id
from pages
group by url
,visitor_id
,duration) as data
where data.avg_by_visitor_id<=data.duration