Difference two rows in a single SQL SELECT statement

Question

I have a database table that has a structure like the one shown below:

CREATE TABLE dated_records (
              recdate DATE NOT NULL
              col1    DOUBLE NOT NULL,
              col2    DOUBLE NOT NULL,
              col3    DOUBLE NOT NULL,
              col4    DOUBLE NOT NULL,
              col5    DOUBLE NOT NULL,
              col6    DOUBLE NOT NULL,
              col7    DOUBLE NOT NULL,
              col8    DOUBLE NOT NULL
              );

I want to write an SQL statement that will allow me to return a record containing the changes between two supplied dates, for specified columns - eg col1, col2 and col3

for example, if I wanted to see how much the value in col1, col2 and col3 has changed during the interval between two dates. A dumb way of doing this would be to select the rows (separately) for each date and then difference the fields outside the db server -

SQL1 = "SELECT col1, col2 col3 FROM dated_records WHERE recdate='2001-01-01'";
SQL1 = "SELECT col1, col2 col3 FROM dated_records WHERE recdate='2001-02-01'";

however, I'm sure there there is a way a smarter way of performing the differencing using pure SQL. I am guessing that it will involve using a self join (and possibly a nested subquery), but I may be over complicating things - I decided it would be better to ask the SQL experts on here to see how they would solve this problem in the most efficient way.

Ideally the SQL should be DB agnostic, but if it needs to be tied to be a particular db, then it would have to be PostgreSQL.

Answer 1

Just select the two rows, join them into one, and subtract the values:

select d1.recdate, d2.recdate,
       (d2.col1 - d1.col1) as delta_col1,
       (d2.col2 - d1.col2) as delta_col2,
       ...
from (select *
      from dated_records
      where recdate = <date1>
     ) d1 cross join
     (select *
      from dated_records
      where recdate = <date2>
     ) d2

Answer 2

I think that if what you want to do is get in the result set rows that doesn't intersect the two select queries , you can use the EXCEPT operator :

The EXCEPT operator returns the rows that are in the first result set but not in the second.

So your two queries will become one single query with the except operator joining them :

SELECT col1, col2 col3 FROM dated_records WHERE recdate='2001-01-01'
EXCEPT
SELECT col1, col2 col3 FROM dated_records WHERE recdate='2001-02-01'

Answer 3

SELECT
COALESCE
(a.col1 -
  (
    SELECT b.col1
    FROM dated_records b
    WHERE b.id = a.id + 1
  ),
a.col1)
FROM dated_records a
WHERE recdate='2001-01-01';

Answer 4

You could use window functions plus DISTINCT :

SELECT DISTINCT
       first_value(recdate) OVER () AS date1
      ,last_value(recdate)  OVER () AS date2
      ,last_value(col1)     OVER () - first_value(col1) OVER () AS delta1
      ,last_value(col2)     OVER () - first_value(col2) OVER () AS delta2
       ...
FROM   dated_records
WHERE  recdate IN ('2001-01-01', '2001-01-03')

For any two days. Uses a single index or table scan, so it should be fast.

I did not order the window, but all calculations use the same window, so the values are consistent.

This solution can easily be generalized for calculations between n rows. You may want to use nth_value() from the Postgres arsenal of window functions in this case.

Answer 5

This seemed a quicker way to write this if you are looking for a simple delta.

SELECT first(col1) - last(col1) AS delta_col1
, first(col2) - last(col2) AS delta_col2
FROM dated_records WHERE recdate IN ('2001-02-01', '2001-01-01')

You may not know whether the first row or the second row comes first, but you can always wrap the answer in abs(first(col1)-last(col1))