问题描述
我有表t1，我想将其与下面的表t2的a，b和c列连接
+---------+---------+---------+
|a |b |c |
+---------+---------+---------+
|473200 |1 |1.-1-1 |
|472400 |10 |1.-1-1 |
|472800 |10 |1.-1-1 |
|473200 |93 |1.-1-1 |
|472800 |26240 |1.-1-1 |
+---------+---------+---------+
t2
+---------+---------+---------+
|a |b |c |
+---------+---------+---------+
|473200 |1 |1.-1-1 |
|472400 |10 |1.-1-1 |
|472800 |10 |1.-1-1 |
|473200 |93 |1.-1-1 |
|472800 |26250 |1.-1-1 |
+---------+---------+---------+
当我仅加入a和c时，结果是
+---------+---------+---------+---------+
|t1.b |t2.b |a |c |
+---------+---------+---------+---------+
|93 |1 |473200 |1.-1-1 |
|1 |1 |473200 |1.-1-1 |
|10 |10 |472400 |1.-1-1 |
|10 |10 |472800 |1.-1-1 |
|26240 |10 |472800 |1.-1-1 |
|93 |93 |473200 |1.-1-1 |
|1 |93 |473200 |1.-1-1 |
|10 |26250 |472800 |1.-1-1 |
|26240 |26250 |472800 |1.-1-1 |
+---------+---------+---------+---------+
我想要实现的是将b列添加到'on'子句中，以便在b列的最小差异处进行连接。
所需结果
+---------+---------+---------+---------+
|t1.b |t2.b |a |c |
+---------+---------+---------+---------+
|1 |1 |473200 |1.-1-1 |
|10 |10 |472400 |1.-1-1 |
|10 |10 |472800 |1.-1-1 |
|93 |93 |473200 |1.-1-1 |
|26240 |26250 |472800 |1.-1-1 |
+---------+---------+---------+---------+
我在这里看到了类似的东西
https://dba.stackexchange.com/questions/73804/how-to-retrieve-closest-value-based-on-look-up-table
但不确定如何适用于我的案件。
解决方法
一种选择是横向连接：
select t1.*,t2.b b2
from t1
cross join lateral (
select t2.*
from t2
where t2.a = t1.a and t2.c = t1.c
order by abs(t2.b - t1.b)
limit 1
)
另一种可能性是distinct on-但您需要t1的主键。假设(a,c)元组唯一地标识t1中的每一行，您将执行以下操作：
select distinct on (t1.a,t1.c) t1.*,t2.b b2
from t1
inner join t2 on t2.a = t1.a and t2.c = t1.c
order by t1.a,t1.c,abs(t2.b - t1.b)
加入表格并计算列c的差异，然后使用distinct on来按差异排序的(a,c)仅返回一行。
with joined as (
select t1.a,t1.b as b1,t2.b as b2,t2.b - t1.b as b_diff
from t1
join t2
on t2.a = t1.a
and t2.b = t1.b
and t1.b <= t2.b
)
select distinct on (a,c) b1,b2,a,c
from joined
order by a,c,b_diff
;