Returning Minimum Rows For Each ID SQL

Question

I have a very long list that looks roughlt like the below (column names and data anonymised)

ID | ID2 | Value
1  |  5  |  900
1  |  7  |  400
2  |  2  |  100
2  |  8  |  800
2  |  2  |  200
3  |  4  |  100
3  |  5  |  300
4  |  8  |  750
4  |  5  |  900

And what I am after is the ability to make 1 row per column "ID", which has the sum of the values of the lowest ID2.

As an example, the above table would turn into:

ID | ID2 | Value
1  |  5  |  900
2  |  2  |  300
3  |  4  |  100
4  |  5  |  900

I have tried several script variations to this with no success. This is my current cleaned up code:

    SELECT res.id
       , min(id2) as rdk
       ,sum(value)

  FROM x as res

    left join y as clai
    on res.id = clai.id

    left join z as cal
    on clai.col99 = cal.col99

    group by
       res.id

However the sum is adding up everything in value for the "ID", not just the lowest "ID2"?

Any help appreciated!

@KobyDouek it would be 300 as there are two entries with an ID = 2, and an ID2 = 2 so it sums the two values together, if that makes sense? — Mike Langlois, May 05 '17 at 08:44

score 1 · Answer 1 · answered May 05 '17 at 08:47

1

Try this:-

select a.ID,a.ID2, sum(Value) as Value
from
your_table_name a
inner join
(
SElect ID, min(ID2) as ID2
from your_table_name
group by ID
) b
on a.ID=b.ID and a.ID2=b.ID2
group by a.ID,a.ID2;

Hope this helps:-)

answered May 05 '17 at 08:47

India.Rocket

1,225
1
8
11

Thanks very much for your help :) – Mike Langlois May 05 '17 at 10:27

score 0 · Accepted Answer · answered May 05 '17 at 08:55

0

Try this out:

select a.*,b.value from
(select id,min(id2) as id2 
from have
group by id) a
left join
(select id,id2,sum(value) as value
from have
group by id,id2) b
on a.id=b.id and a.id2=b.id2;

My Output:

id  |id2 |value
1   |5   |900
2   |2   |300
3   |4   |100
4   |5   |900

Please let me know in case of any queries.

answered May 05 '17 at 08:55

G.Arima

1,171
1
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Thanks very much for your help :) – Mike Langlois May 05 '17 at 10:27
Happy to help :) – G.Arima May 05 '17 at 10:52

score 0 · Answer 3 · answered May 05 '17 at 09:04

I'll just leave my solution here:)

with ExampleData as (
  select 1 ID1,  5 ID2,   900 Value union all
  select 1 ID1,  7  ID2,  400 Value union all
  select 2 ID1,  2  ID2,  100 Value union all
  select 2 ID1,  8  ID2,  800 Value union all
  select 2 ID1,  2  ID2,  200 Value union all
  select 3 ID1,  4  ID2,  100 Value union all
  select 3 ID1,  5  ID2,  300 Value union all
  select 4 ID1,  8  ID2,  750 Value union all
  select 4 ID1,  5  ID2,  900 Value)

select ID1,
       ID2, 
       Value 
from (
   select ID1,
          ID2,
          SUM(VALUE) VALUE,
          ROW_NUMBER() OVER (partition by ID1 order by ID2) R 
   from ExampleData
   group by ID1,ID2
) T where T.R=1

Thanks very much for your help :) – Mike Langlois May 05 '17 at 10:27 — Mike Langlois, May 05 '17 at 10:27

score 0 · Answer 4 · answered May 05 '17 at 09:19

0

If your table is called t1:

SELECT t1.id, t1.id2, SUM(t1.value) 
FROM t1 INNER JOIN
(
 SELECT id, min(id2) minId 
 FROM t1
 GROUP BY id
) A
ON A.id = t1.id AND A.minId = t1.id2
GROUP BY t1.id, t1.id2

answered May 05 '17 at 09:19

NP3

1,114
1
8
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Thanks very much for your help :) – Mike Langlois May 05 '17 at 10:27

Returning Minimum Rows For Each ID SQL

4 Answers4