如何在 oracle 中找到具有日期数据类型的特定列的平均值？ 我希望平均值为 NUMBER 数据类型

Question

Let say i have this table and i have inserted alot of date into it.

CREATE TABLE dateOfTransaction(
    TransDATE  DATE  NOT NULL);

I need to convert the Transdate to number datatype first? then do

SELECT AVG(TransDate) FROM dateOfTransaction;

Answer 1

I don't quite understand what that "average date" is, but - just for fun - have a look at his:

Set date format (to see what is what):

SQL> alter session set nls_date_format = 'dd.mm.yyyy hh24:mi:ss';

Session altered.

Sample data:

SQL> create table dateoftransaction
  2    (transdate date);

Table created.

SQL> insert into dateoftransaction
  2    select sysdate from dual union all
  3    select date '2018-11-23' from dual union all
  4    select to_date('30.09.2025 13:32:33', 'dd.mm.yyyy hh24:mi:ss') from dual;

3 rows created.

SQL> select * from dateoftransaction;

TRANSDATE
-------------------
13.11.2019 20:59:00
23.11.2018 00:00:00
30.09.2025 13:32:33

SQL>

Average date is - of course - invalid:

SQL> select avg(transdate) from dateoftransaction;
select avg(transdate) from dateoftransaction
           *
ERROR at line 1:
ORA-00932: inconsistent datatypes: expected NUMBER got DATE

But, what if we convert it to a Julian date?

SQL> select transdate, to_char(transdate, 'J') julian_date from dateoftransaction;

TRANSDATE           JULIAN_
------------------- -------
13.11.2019 20:59:00 2458801
23.11.2018 00:00:00 2458446
30.09.2025 13:32:33 2460949

That is a number, and yes - we know how to calculate average value on numbers, right?

SQL> select
  2                  avg(to_number(to_char(transdate, 'J')))          avg_julian_date,
  3    to_date(trunc(avg(to_number(to_char(transdate, 'J')))), 'JSP') avg_date
  4  from dateoftransaction;

AVG_JULIAN_DATE AVG_DATE
--------------- -------------------
     2459398,67 02.07.2021 00:00:00

See if it helps.

Answer 2

Subtract a base date, take the average, and add back to the base:

select date '2000-01-01' + avg(dateOfTransaction - date '2000-01-01') * interval '1 day'