How to extract the Datetime from string?

Question

Hello FOlks,
I have the sample data whichi is a string and it contains DateTime part inside it. I worked through half way but am not sure how to get the other part.
The sample data is
Note Added by AARON HILTON at 05/26/2010 14:53 MONEY NOTE ENTERED
Note Added by DEBBIE BILL at 11/16/2009 23:12 MONEY ENTERED
I tried using
substr(d.assignee_note, INSTR(d.assignee_note,'at')+2,length(d.assignee_note))  but it did not have the desired results.
How to just extract the datetime part i.e. 11/16/2009 23:12

Answer 1

If your sample data is representative for all the data, you could also try a regular expression:
SQL> -- generating sample data, pretending it's stored in a column called str in a table called t:
SQL> with t as (
  2  select 'Note Added by AARON HILTON at 05/26/2010 14:53 MONEY NOTE ENTERED' str from dual union
  3  select 'Note Added by DEBBIE BILL at 11/16/2009 23:12 MONEY ENTERED' from dual
  4  )
  5  --
  6  -- actual query:
  7  --
  8  select trim(regexp_replace(str, '[[:alpha:]]')) str
  9  ,      to_date(trim(regexp_replace(str, '[[:alpha:]]')), 'mm-dd-yyyy hh24:mi') dt
10  ,      to_timestamp(trim(regexp_replace(str, '[[:alpha:]]')), 'mm-dd-yyyy hh24:mi') ts
11  from   t;
STR                DT                        TS
05/26/2010 14:53   26-05-2010 14:53:00       26-05-10 14:53:00.000000000
11/16/2009 23:12   16-11-2009 23:12:00       16-11-09 23:12:00.000000000
2 rows selected.where column STR is, well, a string, DT will turn your string into date datatype and TS will turn it into a timestamp.
It isn't clear how you want to extract the datetime part, but it really matters...