如何在 oracle 中基于今年的 7 天获得 2018 年的 1 周数据？

Question

I am trying to dynamically retrieve data based on dates. For example I want data from 7 days before today and same 7 days from 2020 dynamically.

I tried

SELECT *
  FROM table_1
 WHERE  insert_date > TRUNC(SYSDATE) - 7 or ((insert_date< trunc(sysdate)-365) and (insert_date> trunc(sysdate)-372))
 order by insert_date

The problem with this query is if I were to run this query now, this will give me correct data for 2020 and 2021. However if I were to run this same query in 2022, it will give me data from 2022 and 2021 when I want is data based on 2022 and 2020. I was hoping if someone could help me with this issue. I was able to figure out if I want to compare month to month but not week..

Thank you,

Sam

Answer 1

Presume that

• today is (yyyy-mm-dd) 2021-04-08 which means that you'd want to return rows whose insert_date is

  • if you run the query "today", between
    • 2021-04-01 and 2021-04-08
    • 2020-04-01 and 2020-04-08
  • if you run the query "today next year" (ie 2022-04-08 )
    • 2022-04-01 and 2022-04-08
    • 2020-04-01 and 2020-04-08

• this is contents of your table:

   SQL> select * from table_1 order by id; ID INSERT_DAT Return in ---------- ---------- 1 2021-04-08 -- 2021 2 2021-04-03 -- 2021 3 2021-04-02 -- 2021 4 2021-03-31 5 2020-04-05 -- 2021 and 2022 6 2020-04-04 -- 2021 and 2022 7 2020-04-12 8 2022-04-08 -- 2022 9 2022-03-30 9 rows selected.

Query; lines #4 and 5 make sure that query returns rows in year 2020:

SQL> select sysdate from dual;

SYSDATE
----------
2021-04-08

SQL> select a.*
  2  from table_1 a
  3  where a.insert_date between trunc(sysdate) - 7 and trunc(sysdate)
  4     or a.insert_date between add_months(trunc(sysdate), -12 * (extract(year from sysdate) - 2020)) - 7
  5                          and add_months(trunc(sysdate), -12 * (extract(year from sysdate) - 2020))
  6  order by a.insert_date desc;

        ID INSERT_DAT
---------- ----------
         1 2021-04-08
         2 2021-04-03
         3 2021-04-02
         5 2020-04-05
         6 2020-04-04

SQL>

Next year, on 2022-04-08, query would return

SQL> select sysdate from dual;

SYSDATE
----------
2022-04-08

SQL> select a.*
  2  from table_1 a
  3  where a.insert_date between trunc(sysdate) - 7 and trunc(sysdate)
  4     or a.insert_date between add_months(trunc(sysdate), -12 * (extract(year from sysdate) - 2020)) - 7
  5                          and add_months(trunc(sysdate), -12 * (extract(year from sysdate) - 2020))
  6  order by a.insert_date desc;

        ID INSERT_DAT
---------- ----------
         8 2022-04-08
         5 2020-04-05
         6 2020-04-04

SQL>

Answer 2

Didn't quite got the requirement. But, assuming you want to take 2020 as a standard date and get 7 days of data from today and the same 7 days from 2020.

I had used the substr to remove the year part and replace it with the standard year. As I had difficulty testing with 2022 date, took 2019 as my standard date.

 select distinct cast(to_date(p.insert_date, 'DD-MON-YYYY') as timestamp) as insert_date, 
 cast(to_date(SYSDATE-7, 'DD-MON-YYYY')   as timestamp) as tst_date
 from table_1 p
    where 1=1 and  ( (to_date(p.insert_date, 'DD-MON-YYYY') > to_date(SYSDATE-7, 'DD-MON-YYYY') )
                  or (to_date(p.insert_date, 'DD-MON-YYYY')> trunc(to_date(substr(to_date(SYSDATE, 'DD-MON-YYYY'),1,7)||'19', 'DD-MON-YYYY')-7)
                        and to_date(p.insert_date, 'DD-MON-YYYY') <= trunc(to_date(substr(to_date(SYSDATE, 'DD-MON-YYYY'),1,7)||'19', 'DD-MON-YYYY')))
                  )
     order by create_date desc; 

  -- Standard date for 2020
  select substr(to_date(SYSDATE, 'DD-MON-YYYY'),1,7)||'20' from dual;