CODEWITHSUNDEEP
mysqlaverage
Lorenzo W.
Lorenzo W.

Reputation: 27

Average number of days using MySQL

I'm trying to get the average number of days between when a user closes a ticket to when they open their "next" ticket.

A sample of my MySQL table is below. You will notice that Mike has created 3 tickets, 9 days apart (July1-July 10) and 5 days (July 10-July 15) apart, average is 7 days. I can't seem to figure out how to look for the last resolved date, anybody have any ideas? This is what I have so far:

SELECT 
  Name, 
  Created, 
  Resolved, 
  avg(datediff("Last Ticket Resolved", created) AS last_tket_open 
FROM 
  MyTable 
WHERE 
  Name='Mike'


Name   Created  Resolved
----   -------  --------
Mike   July  1   July  1
Jill   July  2   July  3    
Mike   July 10   July 10
Harry  July 11   July 11
Mike   July 15   July 15

Upvotes: 0

Views: 1552

Answers (3)

SetFreeByTruth
SetFreeByTruth

Reputation: 847

Try this:

SELECT
  t1.Name,
  AVG( datediff( t2.Created, t1.Resolved )) AS avg_days_between_tickets
FROM
  MyTable t1,
  MyTable t2
WHERE
      t2.Name = t1.Name
  AND t2.Created = ( SELECT MIN( Created ) 
                     FROM MyTable 
                     WHERE Name = t1.Name 
                       AND Created > t1.Resolved )
GROUP BY
  t1.Name

Upvotes: 0

Zane Bien
Zane Bien

Reputation: 23125

SELECT a.name, 
       AVG(DATEDIFF(a.resolved, b.created)) AS avgdays
FROM
(
   SELECT name, resolved, @val1:=@val1+1 AS rn
   FROM tbl
   CROSS JOIN (SELECT @val1:=0) val1_init
   WHERE name = 'Mike'
   ORDER BY resolved
) a
INNER JOIN
(
   SELECT created, @val2:=@val2+1 AS rn
   FROM tbl
   CROSS JOIN (SELECT @val2:=1) val2_init
   WHERE name = 'Mike'
   ORDER BY resolved
) b ON a.rn = b.rn

SQLFiddle Demo

Upvotes: 1

Bohemian
Bohemian

Reputation: 425043

select 
  Name,
  AVG( datediff( next_created, resolve )) AS avg_time
from (select
  t1.Name,
  t1.Created,
  t1.Resolved,
  min(t2.Created) as next_created
from mytable t1
join myTable t2 on t1.Name = t2.Name and t2.Created > t1.Resolved
group by 1,2,3) x
group by 1;

Note that this query does not use an correlated subqueries (queries that execute for every row due to the way they are coded to depend on a value in a row), so it should perform quite well.

Upvotes: 0

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