PostgreSQL query with max and min date plus associated id per row
You can calculate this in one step with window functions:

CREATE OR REPLACE VIEW daily_trans AS
SELECT DISTINCT
       trans_date
     , first_value(trans_time) OVER w AS first_time
     , first_value(id)         OVER w AS first_id
     , last_value(trans_time)  OVER w AS last_time
     , last_value(id)          OVER w AS last_id
     , calculate_status(min(trans_time) OVER w
                      , max(trans_time) OVER w) AS status 
FROM   trans
WINDOW w AS (PARTITION BY trans_date ORDER BY trans_time, id
             ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)
ORDER  BY 1;


I added id as secondary ORDER column in the window clause, to make the sort order stable in case of identical times per day.

If you are not familiar with window functions, be sure to read this chapter of the manual.

ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING is a not so commonly used clause to adjust the window frame, because we want last_value() to always return the last value of the whole partition (the date), not the last value up to the current (including peers) according to the sort order.

How you to combine DISTINCT with window functions:


PostgreSQL: running count of rows for a query 'by minute'