MySQL组连续出场

时间:2010-08-13 12:28:25

标签: mysql group-by

我想通过连续出现列值来对查询结果进行分组。假设我有一张表格列出了每年比赛的获胜者,如下所示:

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

我想查询输出:

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

我不太担心“start_end”的格式,因为我有开始和结束或范围(例如,可以使用GROUP_CONCAT来生成2004,2005,2006而不是2004-2006,那就是还可以。)

1 个答案:

答案 0 :(得分:3)

如果您的表格如下:

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

此查询应该可以解决问题:

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你可能会发现Common Queries Tree解决这些问题很方便(查看“查找序列中的上一个和下一个值”的答案):p。