MySQL分组连续出现
作者:互联网
我想按列值的连续出现将查询结果分组.假设我有一张表格,其中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:
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,我就不用太担心“ start_end”的格式,那仍然可以) ).
解决方法:
只要您的表格如下所示:
"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这个问题:
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;
顺便说一句,您可能会发现Common Queries Tree可以轻松解决这些问题(请查看“查找序列中的上一个和下一个值”的答案):p.
标签:mysql,group-by 来源: https://codeday.me/bug/20191209/2097250.html