SQL Server: select from table with/without aggregate columns using groupby

Question

I have the columns and rows for the table A,

tableid	featureid	col3	col4	coldatetime
1	1	AD	4	2022-06-22 09:00:00
2	2	BC	5	2022-06-22 09:00:00
3	1	AE	6	2022-06-22 10:00:00
4	3	BD	7	2022-06-22 11:00:00
5	2	BB	8	2022-06-22 16:00:00

I need the following result in the SQL Server,

featureid	col3	col4
1	AD	4,6
2	BC	5,8
3	BD	7

If I run the following query:

select featureid,
       STRING_AGG(col3,',') as col3,
       STRING_AGG(col4,',') as col4 
from table_server 
group by feature_id;

I am getting the following result,

featureid	col3	col4
1	AD,AE	4,6
2	BC,BD	5,8
3	BD	7

How Should I change my query to get the Col3 to have only one record?

I have this clarification, whether this is possible or not?

I have zero knowledge with SQL Server, can anyone help me?

Answer 1

Seems like you need MIN instead of STRING_AGG

select
  s.featureid,
  MIN(s.col3, ',') as col3,
  STRING_AGG(s.col4, ',') as col4 
from table_server s
group by
  s.feature_id;

Answer 2

If you first transform the irrelevant " col3 " values to NULL, you can use your own code, as long as the STRING_AGG function ignores them completely.

WITH cte AS (
    SELECT tableid,
           featureid,
           CASE WHEN col4 = MIN(col4) OVER(PARTITION BY featureid) 
                THEN col3 
           END AS col3,
           col4
    FROM table_server
)
select featureid,
       STRING_AGG(col3,',') as col3,
       STRING_AGG(col4,',') as col4 
from cte 
group by featureid;

Check the demo here .

Note : this solution assumes that the " col3 " value you want to keep is linked to the lowest " col4 " value, with no ordering assumption on the " col3 " field itself.