GROUP BY with MAX and UNION, or JOIN? - sql

Question

How to obtain from this table date_departure and date_arrival for each travel according visiting_order
select * from step;
id_step | id_travel | id_port | visiting_order | date_arrival | date_departure
---------+-----------+---------+----------------+--------------+----------------
1 | 1 | 1 | 0 | | 2021-01-12
2 | 1 | 2 | 1 | 2021-05-20 | 2021-05-22
3 | 1 | 3 | 2 | 2021-07-27 |
4 | 2 | 4 | 0 | | 2021-02-13
5 | 2 | 5 | 1 | 2021-02-27 |
6 | 3 | 7 | 0 | | 2022-01-12
7 | 3 | 6 | 1 | 2022-05-27 |
like this :
id_travel | date_departure | date_arrival
------------+----------------+--------------
1 | 2021-01-12 | 2021-07-27
2 | 2021-02-13 | 2021-02-27
3 | 2022-01-12 | 2022-05-27
?
My first intention was to take both columns and UNION them
(SELECT id_travel, date_departure FROM step WHERE visiting_order = 0
GROUP BY id_travel, date_departure)
UNION
(SELECT A.id AS id_travel, A.arr_date AS date_arrival FROM
(SELECT id_travel, MAX(visiting_order), date_arrival
FROM step GROUP BY id_travel
) AS A(id, ord, arr_date)
);
and first select is ok
id_travel | date_departure
-----------+----------------
1 | 2021-01-12
2 | 2021-02-13
3 | 2022-01-12
but second one return an error
ERROR: column "step.date_arrival" must appear in the GROUP BY clause or be used in an aggregate function

Answer 1

Seems like this can just be:
SELECT id_travel
, min(date_departure) AS date_departure
, max(date_arrival) AS date_arrival
FROM step
GROUP BY 1
ORDER BY 1;
Certainly works with your sample data.

Answer 2

SELECT DISTINCT step.id_travel, sub_query1.date_departure, sub_query.date_arrival
FROM step
INNER JOIN
(SELECT step.id_travel, MAX(step.date_arrival) as date_arrival
FROM step
GROUP BY step.id_travel
)
AS sub_query ON (sub_query.id_travel = step.id_travel)
INNER JOIN
(SELECT id_travel, date_departure FROM step WHERE visiting_order = 0 GROUP BY id_travel, date_departure)
AS sub_query1 ON (sub_query1.id_travel = step.id_travel)
ORDER BY id_travel;