1
l00t 2020-12-29 23:18:37 +08:00
看不懂……
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2
wanv1171 2020-12-29 23:38:35 +08:00
先进行分组,然后用 over 选出每组最小的 B 。然后重新根据这个结果筛选 C IN (SELECT C) AND B IN (SELECT B),然后 GROUP BY C, GROUP_CONCAT(A)
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3
wanv1171 2020-12-29 23:44:36 +08:00 via iPhone
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4
venhal 2020-12-29 23:44:55 +08:00
可以了解一下开窗函数,用 row_number()over()应该就可以满足
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5
venhal 2020-12-29 23:52:46 +08:00
不过低版本的 mysql 好像不支持开窗函数,如果不支持的话
select t1.a, t2.b, t2.c from t1 join (select c, min(b) as b from t group by c) as t2 ON t1.c = t2.c AND t1.b = t2.b |
6
lybcyd 2020-12-30 11:22:10 +08:00 via Android
子查询吧,先用 c 分组,用 min 函数查出最小的 b 值,作为一个表 x,然后再查询大表中 b 和 c 等于 x 中 b 和 c 的记录
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8
liquorppp OP |
9
l00t 2020-12-30 12:26:15 +08:00
两个售价一样低的你打算怎么整。
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10
weiwenhao 2020-12-30 14:20:19 +08:00
select * from logs INNER JOIN (select MIN(id) as min_id FROM logs GROUP BY user_id) test on `test`.min_id = logs.id
类似这样? |
11
ZanderNg 2020-12-30 14:34:34 +08:00
select t1.*
from (select t.a, t.b, t.c, row_number() over(partition by t.a order by to_number(b)) as rank from tgxt.test_a t where t.c is not null) t1 where rank = 1 |
14
xdwmxx 2020-12-30 20:35:20 +08:00
select *
from (select t.A, t.B, t.C, row_number() over(partition by t.C order by t.B) rn from TEST_ROW_NUMBER_OVER t) where t.B is not null and rn = 1; 这样吗? |