select distinct list of ids from table with earliest value in same table - sql

Question

I have the following table,
SDate Id Balance
2016-01-01 ABC 3
2016-01-01 DEF 7
2016-01-01 GHI 2
2016-02-01 ABC 6
2016-02-01 DEF 4
2016-02-01 GHI 8
2016-02-01 XYZ 12
I need to write a query that gives me a distinct list of Id's over a date range (so in this example SDate >= '2016-01-01' and SDate <= '2016-02-01') but also give me the earliest balance so the result from the table above I would like to see is,
Id Balance
ABC 3
DEF 7
GHI 2
XYZ 12
Is this possible?
UPDATE
Sorry I should have specified that for each date the Id is unique.

Answer 1

You can do this with a derived table that first works out the minimum SDate value for each Id value. Using this you then join back to your original table to find the Balance for the row that matches those values:
declare #t table(SDate date,Id nvarchar(3),Balance int);
insert into #t values ('2016-01-01','ABC',3),('2016-01-01','DEF',7),('2016-01-01','GHI',2),('2016-02-01','ABC',6),('2016-02-01','DEF',4),('2016-02-01','GHI',8),('2016-02-01','XYZ',12);
declare #StartDate date = '20160101';
declare #EndDate date = '20160201';
with d as
(
select Id
,min(SDate) as MinSDate
from #t
where SDate between #StartDate and #EndDate
group by id
)
select d.Id
,t.Balance
from d
inner join #t t
on(d.Id = t.Id
and d.MinSDate = t.SDate
);
Output:
Id | Balance
----+--------
ABC | 3
DEF | 7
GHI | 2
XYZ | 12

Answer 2

This should be possible with a window function - all you have to do is
partition by id
assign a row number, and
select the top row for each id
Example:
select id,
balance
from (
select id,
balance,
row_number() over( partition by id order by SDate ) as row_num
from table1
where SDate between '2016-01-01' and '2016-02-01'
) as a
where row_num = 1
Note: the advantage of this method is it is a lot more flexible. Say you wanted the 2 oldest records, you could just change to where row_num <= 2.

Answer 3

Analytic row_number() should be the fastest
select *
from (
select
t.*,
row_number() over (partition by Id order by SDate) rn
from your_table t
) t where rn = 1;

Answer 4

You can achieve this with a self join, which may not be the fastest or most elegant solution:
CREATE TABLE #SOPostSample
(
SDate DATE ,
Id NVARCHAR(5) ,
Balance INT
);
INSERT INTO #SOPostSample
( SDate, Id, Balance )
VALUES ( '2016-01-01', 'ABC', 3 ),
( '2016-01-01', 'DEF', 7 ),
( '2016-01-01', 'GHI', 2 ),
( '2016-02-01', 'ABC', 6 ),
( '2016-02-01', 'DEF', 4 ),
( '2016-02-01', 'GHI', 8 ),
( '2016-02-01', 'XYZ', 12 );
SELECT t1.Id ,
MIN(t2.Balance) Balance
FROM #SOPostSample t1
INNER JOIN #SOPostSample t2 ON t1.Id = t2.Id
GROUP BY t1.Id ,
t2.SDate
HAVING t2.SDate = MIN(t1.SDate);
DROP TABLE #SOPostSample;
Produces:
id Balance
============
ABC 3
DEF 7
GHI 2
XYZ 12
This works for the sample data, but please test with more data as I just wrote it quickly.

Answer 5

This should work, Top 1 just inserted for safety, should not be needed if SDate and Id are unique in combination
SELECT o.Id ,
( SELECT TOP 1
Balance
FROM tbl
WHERE Id = o.Id
AND SDate = MIN(o.SDate)
) Balance
FROM tbl o
GROUP BY Id
HAVING sDate BETWEEN '20160101' AND '20160201';

Answer 6

You can use sub-query
SELECT Id ,
( SELECT TOP 1
Balance
FROM [TableName] AS T1
WHERE T1.Id = [TableName].Id
ORDER BY SDate
) AS Balance
FROM [TableName]
GROUP BY Id;