Find first date missing in postgresql. Have a table and I want to find out first non-existing date starting from now() to now()+30days. Lets assume today is 21.

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i have a table and i want to find out first non-existing date starting from now() to now()+30days. lets assume today is 21.

>| userId |       dates        |
| 1      | 2021-06-22 00:00:00|
| 1      | 2021-06-24 00:00:00|

| 2      | 2021-06-21 00:00:00|
| 2      | 2021-06-22 00:00:00|
| 2      | 2021-06-23 00:00:00|
| 2      | 2021-06-30 00:00:00|

| 3      | 2021-06-23 00:00:00|
| 3      | 2021-06-24 00:00:00|
| 3      | 2021-06-27 00:00:00|
expected output :-

| userId |       dates        |
| 1      | 2021-06-21 00:00:00|
| 2      | 2021-06-24 00:00:00|
| 3      | 2021-06-21 00:00:00|
I'm using postgres. can u help? the data is pretty large

What I have tried:

used lead() but didn't work and taking max min also didn't work.

Posted 21-Jun-21 10:31am

a_horse_with_no_name_2

Updated 21-Jun-21 21:10pm

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Maciej Los · Accepted Answer · 2021-06-21T21:10:00

You can use common table expressions for such of requirements. See:

SQL

WITH RECURSIVE alldays as
(
  -- initial query
  SELECT CAST('2021-06-21' as timestamp without time zone) currdate, 
    CAST('2021-06-21' as timestamp without time zone) + 30 * interval '1 day' maxdate
  -- recursive part
  UNION ALL
  SELECT a.currdate + interval '1 day' currdate, a.maxdate
  FROM alldays a
  WHERE a.currdate + interval '1 day' <= a.maxdate
),
allusers as
(
  SELECT DISTINCT userid
  FROM mydata
)
SELECT u.userid, d.currdate dates
FROM allusers u CROSS JOIN alldays d
WHERE NOT EXISTS
(
  SELECT userid, dates
  FROM mydata md
  WHERE md.userid = u.userid AND md.dates = d.currdate
)
ORDER BY u.userid, d.currdate;

dbfiddle[^]

Note:
Above query returns all missing dates till now + 30 days. Change it to your needs.