Finding closest value in another table - sql

Question

I'm quite new to SQL so bare with me. What I'm trying to do is return the value closest to another value in a different table for every record.
I'll show a simplified example of my two tables for clarification
First table is the one that I want the value ENTRY_YEAR matched to:
ID
ENTRY_VALUE
1001
1900
1002
2000
And the second table:
ID
ENTRY_VALUE
STATUS
1001
1880
SUCCES
1001
1930
FAIL
1001
1940
SUCCES
1002
1960
SUCCES
1002
1980
FAIL
So the end result I'm looking for is:
ID
ENTRY_VALUE
STATUS
1001
1880
SUCCES
1002
1980
FAIL
I have currently only managed to link the id's together but can't find a way to compare the ENTRY_VALUE in both tables and return the one closest to the Table1 entry.
So only this:
SELECT * from Table2
INNER JOIN Table1 ON (Table2.ID = Table1.ID)
Once again my bad for the basic question, I have googled right about everything but can't get it to work so any help is very welcome!

Answer 1

First attempt
This is a (slower performing) query. First attempt! This is an approach using a "correlated subquery" so it runs the inner query for each row of the outer query. The strategy is, for each row, to determine what the min value is we are looking for, and then select only the rows that fit that criteria. But such queries can be slow at runtime, although the logic is very clean.
select
a.id,
b.entry_value,
b.[status]
from
Foo a
inner join Bar b
on a.id = b.id
where
abs(a.entry_value - b.entry_value) =
(select min(abs(t1.entry_value-t2.entry_value))
from Foo t1
inner join Bar t2
on t1.id = t2.id
where
t1.id = a.id
group by t1.id)
Second attempt
If you have many rows (in the tens of thousands or in any case if the previous query is just too slow), then this next one should be better performing. Second Attempt! If you run the two inner queries by themselves, you will probably see the strategy here of how we are joining them to get the desired result.
select A.Id, A.entry_value, A.[status]
from
(
select t1.id, t2.entry_value, abs(t1.entry_value-t2.entry_value) as diff, t2.[status]
from Foo t1
inner join Bar t2
on t1.id = t2.id
) A
inner join
(
select t3.id, min(abs(t3.entry_value-t4.entry_value)) as diff
from Foo t3
inner join Bar t4
on t3.id = t4.id
group by t3.id
) B
on A.id = B.id
and A.diff = B.diff
Note
I would probably not try to write either of these queries in MSAccess "Design view" although if I had too I am sure I could. But generally, this is a case where I would write the query "by hand" and paste it into your query directly using MSAccess "SQL view".
Caution
Beware that ties will result in two rows! Example:
First table has (1003,2000)
Second table has (1003, 1990, 'success') and (1003, 2010, 'fail')
You will have a result with two rows, one with success and the other with fail (!)
So you really should test with your data and look for such cases that might produce such ties (and decide what to do, if necessary).
Btw...
just for fun, here's how you might go for it in SQL Server.
But I think this will NOT work in MSAccess, unfortunately.
select
T.id,
T.entry_value,
T.[status]
from
(
select
t1.id,
t2.entry_value,
abs(t1.entry_value-t2.entry_value) as diff,
t2.[status],
rank() over (partition by t1.id order by abs(t1.entry_value-t2.entry_value)) as seq
from #Foo t1
inner join #Bar t2
on t1.id = t2.id
) T
where T.seq = 1;

Answer 2

Use a simple subquery to find the minimum offset:
Select
tbl1.ID,
tbl2.ENTRY_VALUE,
tbl2.STATUS
From
tbl1
Inner Join
tbl2 On tbl1.ID = tbl2.ID
Where
Abs([tbl1].[ENTRY_VALUE] - [tbl2].[ENTRY_VALUE]) =
(Select Min(Abs([tbl1].[ENTRY_VALUE] - [T2].[ENTRY_VALUE])) As Offset
From tbl2 As T2
Where T2.ID = tbl1.ID);
Output:
ID
ENTRY_VALUE
STATUS
1001
1880
SUCCES
1002
1980
FAIL
Note, that if the minimum offset for an ID exists twice, both records having this offset will be returned. Thus, you may have to aggregate the output.