Oracle SQL - Column with unix timestamp, need dd-mm-yyyy timestamp

Question

Hi,

is there any way in Oracle, to get only the dd-mm-yyyy part from an unix timestamp in oracle?

Like:

select to_char(my_timestamp, 'ddmmyyyy') as my_new_timestamp from table

Answer 1

I believe it's:

select to_char(my_timestamp, 'dd-mm-yyyy') as my_new_timestamp from table

See also this reference on Oracle's date format specifiers.

Answer 2

As I understand it, A Unix timestamp is defined as a number of seconds since 1970-01-01, in which case:

select DATE '1970-01-01' + my_timestamp/86400 from table;

(There are 86400 seconds in a day.)

To ignore the hours, minutes and seconds:

select TRUNC(DATE '1970-01-01' + my_timestamp/86400) from table;

However, if what you want is a "truncated" Unix timestamp then try this:

select floor(my_timestamp/84600)*84600 from dual;

Answer 3

http://forums.devshed.com/oracle-development-96/convert-unix-timestamp-to-oracle-data-format-324204.html

Answer 4

Given this data ...

SQL> alter session set nls_date_format='dd-mon-yyyy hh24:mi:ss'
  2  /

Session altered.

SQL> select * from t23
  2  /

MY_TIMESTAMP
--------------------
08-mar-2010 13:06:02
08-mar-2010 13:06:08
13-mar-1985 13:06:26

SQL> 

.. it is simply a matter of converting the time elapsed since 01-JAN-1970 into seconds:

SQL> select my_timestamp
  2        , (my_timestamp - date '1970-01-01') * 86400 as unix_ts
  3  from t23
  4  /

MY_TIMESTAMP            UNIX_TS
-------------------- ----------
08-mar-2010 13:06:02 1268053562
08-mar-2010 13:06:08 1268053568
13-mar-1985 13:06:26  479567186

SQL>