Postgresql: How To Return Rows With Respect To A Found Row (relative Results)?
Solution 1:
createtable foo (dt date);
insertinto foo values
('2012-12-01'),
('2012-08-01'),
('2012-07-01'),
('2012-06-01'),
('2012-05-01'),
('2012-04-01'),
('2012-03-01'),
('2012-02-01'),
('2012-01-01'),
('1997-01-01'),
('2012-09-01'),
('2012-10-01'),
('2012-11-01'),
('2013-01-01')
;
select dt
from (
(
select dt
from foo
where dt <=current_dateorderby dt desc
limit 4
)
unionall
(
select dt
from foo
where dt >current_dateorderby dt
limit 7
)) s
orderby dt
;
dt
------------2012-03-012012-04-012012-05-012012-06-012012-07-012012-08-012012-09-012012-10-012012-11-012012-12-012013-01-01
(11rows)
Solution 2:
You could use the window function lead():
SELECT dt_lead7 AS dt
FROM (
SELECT *, lead(dt, 7) OVER (ORDERBY dt) AS dt_lead7
FROM foo
) d
WHERE dt <= now()::dateORDERBY dt DESC
LIMIT 11;
Somewhat shorter, but the UNION ALL version will be faster with a suitable index.
That leaves a corner case where "date closest to today" is within the first 7 rows. You can pad the original data with 7 rows of -infinity to take care of this:
SELECT d.dt_lead7 AS dt
FROM (
SELECT*, lead(dt, 7) OVER (ORDERBY dt) AS dt_lead7
FROM (
SELECT'-infinity'::dateAS dt FROM generate_series(1,7)
UNIONALLSELECT dt FROM foo
) x
) d
WHERE d.dt <= now()::date-- same as: WHERE dt <= now()::date1ORDERBY d.dt_lead7 DESC-- same as: ORDER BY dt DESC 1
LIMIT 11;
I table-qualified the columns in the second query to clarify what happens. See below.
The result will include NULL values if the "date closest to today" is within the last 7 rows of the base table. You can filter those with an additional sub-select if you need to.
To address your doubts about output names versus column names in the comments - consider the following quotes from the manual.
Where to use an output column's name:
An output column's name can be used to refer to the column's value in
ORDER BYandGROUP BYclauses, but not in theWHEREorHAVINGclauses; there you must write out the expression instead.
Bold emphasis mine. WHERE dt <= now()::date references the column d.dt, not the the output column of the same name - thereby working as intended.
If an
ORDER BYexpression is a simple name that matches both an output column name and an input column name,ORDER BYwill interpret it as the output column name. This is the opposite of the choice thatGROUP BYwill make in the same situation. This inconsistency is made to be compatible with the SQL standard.
Bold emphasis mine again. ORDER BY dt DESC in the example references the output column's name - as intended. Anyway, either columns would sort the same. The only difference could be with the NULL values of the corner case. But that falls flat, too, because:
the default behavior is
NULLS LASTwhenASCis specified or implied, andNULLS FIRSTwhenDESCis specified
As the NULL values come after the biggest values, the order is identical either way.
Or, without LIMIT (as per request in comment):
WITH x AS (
SELECT *
, row_number() OVER (ORDERBY dt) AS rn
, first_value(dt) OVER (ORDERBY (dt > '2011-11-02')
, dt DESC) AS dt_nearest
FROM foo
)
, y AS (
SELECT rn AS rn_nearest
FROM x
WHERE dt = dt_nearest
)
SELECT dt
FROM x, y
WHERE rn BETWEEN rn_nearest - 3AND rn_nearest + 7ORDERBY dt;
If performance is important, I would still go with @Clodoaldo's UNION ALL variant. It will be fastest. Database agnostic SQL will only get you so far. Other RDBMS do not have window functions at all, yet (MySQL), or different function names (like first_val instead of first_value). You might just as well replace LIMIT with TOP n (MS SQL) or whatever the local dialect.
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