SQL – How to get a duplicate column based on the value of another column

November 17, 2022

I have a query that return values based on a boolean column: if the id_crsp includes a boolean true AND false, then it is selected.
Values of id_crsp that have only a true or false value are not selected.

From this result, I would like to sort the id_crsp which have duplicates, and select only the one with the oldest date

Database values :

id	idcrsp	date	boolean
1	100	11-2022	true
2	100	07-2022	false
3	200	06-2022	false
4	300	09-2022	true
5	300	08-2022	false
6	400	10-2022	false
7	100	01-2022	false
8	100	02-2022	false

My actual request :

SELECT true_table.* FROM mydb as true_table 
INNER JOIN 
(SELECT * FROM mydb WHERE requalif=TRUE) as false_table 
ON true_table.idcrsp = false_table.idcrsp 
AND true_table.requalif = FALSE;

This return :

id	idcrsp	date	boolean
8	100	02-2022	false
7	100	01-2022	false
2	100	07-2022	false
5	300	08-2022	false

I would like to enrich my request in order to have only two lines:

id 5
id 7 (which has duplicates of id_crsp and which has the oldest date).

Thanks for your help !

>Solution :

using your original query could you just add row_number and partition it by the idcrsp and then order by date descending. then just pick the rows where rn = 1

select * from
(SELECT true_table.*, row_number() over (partition by id_crsp order by date desc) rn  FROM mydb as true_table 
INNER JOIN 
(SELECT * FROM mydb WHERE requalif=TRUE) as false_table 
ON true_table.idcrsp = false_table.idcrsp 
AND true_table.requalif = FALSE)
where rn = 1