MySQL按连续出现分组

Question

MySQL按连续出现分组

3

我希望能够按照列值的连续出现对查询结果进行分组。比如说，我有一个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

我想要一个查询，输出以下内容：

。

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

只要我有开始和结束或范围（例如，可以使用GROUP_CONCAT生成2004、2005、2006，而不是2004-2006，这仍然可以接受），我就不太担心“start_end”的格式。

- Max

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- wimvds · Accepted Answer

假设您的表格长这样：

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

这个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;

顺便提一下，您可能会发现常用查询树在解决这些问题时非常方便（查看“在序列中查找前一个和后一个值”的答案）:p。