MySQL group by consecutive appearances

Question

Hi all,

I would like to group query results by consecutive appearances of a column values. Let's say I have a table which lists the winners of a competition for each year as follows:

year    team_name
2000    AAA
2001    CCC
2002    CCC
2003    BBB
2004    AAA
2005    AAA
2006    AAA

I would like a query which outputs:

start_end    total   team_name
2000         1       AAA
2001-2002    2       CCC
2003         1       BBB
2004-2006    3       AAA

I'm not too much worried about the format of "start_end" at long as I have the start and end or range (.e.g. one could use GROUP_CONCAT to produce 2004,2005,2006 instead of 2004-2006 and that would still be OK).

Answer 1

Provided that your table looks like this :

"id";"year";"team"
"1";"2000";"AAA"
"2";"2001";"CCC"
"3";"2002";"CCC"
"4";"2003";"BBB"
"5";"2004";"AAA"
"6";"2005";"AAA"
"7";"2006";"AAA"

This query should do the trick :

SELECT a.year AS start
     , MIN(c.year) AS end
     , MIN(c.year)-a.year+1 AS total
     , CONCAT_WS('-', a.year, IF(a.year = min(c.year), NULL, min(c.year))) as start_end
     , a.team
  FROM 
     ( SELECT x.year, x.team, COUNT(*) id
         FROM results x
         JOIN results y
           ON y.year <= x.year
        GROUP BY x.id
     ) AS a
  LEFT JOIN 
     ( SELECT x.year, x.team, COUNT(*) id 
         FROM results x
         JOIN results y
           ON y.year <= x.year
        GROUP BY x.id
     ) AS b ON a.id = b.id + 1 AND b.team = a.team
  LEFT JOIN  
     ( SELECT x.year, x.team, COUNT(*) id 
         FROM results x
         JOIN results y
           ON y.year <= x.year
        GROUP BY x.id
     ) AS c ON a.id <= c.id AND c.team = a.team
  LEFT JOIN 
     ( SELECT x.year, x.team, COUNT(*) id 
         FROM results x
         JOIN results y
           ON y.year <= x.year
        GROUP BY x.id
     ) AS d ON c.id = d.id - 1 AND d.team = c.team
WHERE b.id IS NULL AND c.id IS NOT NULL AND d.id IS NULL
GROUP BY start;

BTW You might find the Common Queries Tree handy to solve these problems (check the answers for "Find previous and next values in a sequence") :p.