I got table t1 and i want to join it with table t2 below on columns a, b and c
x
+---------+---------+---------+|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 |+---------+---------+---------+When I join only on a and c the result is
+---------+---------+---------+---------+|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 |+---------+---------+---------+---------+What I try to achieve is to add column b to ‘on’ clause so that join happens on minimum difference in b column.
Desired result
+---------+---------+---------+---------+|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 |+---------+---------+---------+---------+I saw something similar here
https://dba.stackexchange.com/questions/73804/how-to-retrieve-closest-value-based-on-look-up-table
but not sure how to apply to my case.
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Answer
Join the tables and calculate the differences for column c, then use distinct on to return only one row per (a, c) ordered by the difference.
with joined as ( select t1.a, t1.c, 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;