[英]Select distinct values from a column based on highest value in another column
[英]set column value based on distinct values in another column
我试图做一些非常类似于这个问题的东西: mysql - 基于其他行更新行
我有一个名为modset的表,格式如下:
member year y1 y2 y3 y1y2 y2y3 y1y3 y1y2y3
a 1 0 0 0 0 0 0 0
a 2 0 0 0 0 0 0 0
a 3 0 0 0 0 0 0 0
b 1 0 0 0 0 0 0 0
b 2 0 0 0 0 0 0 0
c 1 0 0 0 0 0 0 0
c 3 0 0 0 0 0 0 0
d 2 0 0 0 0 0 0 0
第3:9列是二进制标志,用于指示成员记录的年份组合。因此,我希望SQL更新的结果如下所示:
member year y1 y2 y3 y1y2 y2y3 y1y3 y1y2y3
a 1 0 0 0 0 0 0 1
a 2 0 0 0 0 0 0 1
a 3 0 0 0 0 0 0 1
b 1 0 0 0 1 0 0 0
b 2 0 0 0 1 0 0 0
c 1 0 0 0 0 0 1 0
c 3 0 0 0 0 0 1 0
d 2 0 1 0 0 0 0 0
上面链接的问题中的代码确实非常接近,但只有当它是成员记录的不同年份的计数时。 我需要将列基于成员记录的年份的特定值。
提前致谢!
解
SELECT member,
case when min(distinct(year)) = 1 and max(distinct(year)) = 1 then 1 else 0 end y1,
case when min(distinct(year)) = 1 and max(distinct(year)) = 2 then 1 else 0 end y1y2,
case when min(distinct(year)) = 1 and max(distinct(year)) = 3 and count(distinct(year)) = 2 then 1 else 0 end y1y3,
case when min(distinct(year)) = 1 and max(distinct(year)) = 3 and count(distinct(year)) = 3 then 1 else 0 end y1y2y3,
case when min(distinct(year)) = 2 and max(distinct(year)) = 2 then 1 else 0 end y2,
case when min(distinct(year)) = 2 and max(distinct(year)) = 3 then 1 else 0 end y2y3,
case when min(distinct(year)) = 3 then 1 else 0 end y3
INTO temp5
FROM modset
GROUP BY member;
UPDATE modset M
SET y1 = T.y1, y2 = T.y1, y3 = T.y3, y1y2 = T.y1y2, y1y3 = T.y1y3, y2y3 = T.y2y3, y1y2y3 = T.y1y2y3
FROM temp5 T
WHERE T.member = M.member;
您使用什么查询来返回成员记录的年份指标?
听起来您想要获取查询结果并在更新中使用它:
http://dev.mysql.com/doc/refman/5.0/en/update.html
它可能看起来像这样:
UPDATE targetTable t, sourceTable s
SET t.y1 = s.y1, t.y2 = s.y2 -- (and so on...)
WHERE t.member = s.member AND t.year = m.year;
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