Trying to get a date value from a string

Question

Hi All,
I'm not sure I am doing this right, so I figured I would ask the experts...
I have built a file upload page that is working the way I need it, but I have found that I need to determine the date from the uploaded file name. I tried to create a PL/SQL computation and do some substr / instr to the filename, but I'm not getting it as I get "ORA-01843: not a valid month" error. Here is the code I'm trying:
declare
  v_underscore  integer;
  v_date            varchar2(15);  --have also tried using date datatype...
begin
  v_underscore := INSTR(:P20_EXPRESS,'_',1,1);
  v_date := to_date(substr(:P20_EXPRESS,v_underscore + 1,6),'YYMMDD');
  return v_date;
end;And here is a sample filename:
F15047/proc_10607161001.csv
Is this the best way to achive what I'm trying to do?
Thanks,
Corey

Answer 1

Hi,
v_underscore := INSTR(:P20_EXPRESS,'_',1,1);
v_date := to_date(substr(:P20_EXPRESS,v_underscore + 1,6),'YYMMDD');What this is saying is:
Get the position of the "_" character
Move to the next character to the right (which is a "1" in your case)
Get 6 characters (106071) from the string value
Convert them to a date using the format 'YYMMDD'
This results in:
Year: 10
Month: 60
Day: 71
Which of course is not valid.
For example:
select
  substr('F15047/proc_10607161001.csv', instr('F15047/proc_10607161001.csv', '_', 1, 1) + 1, 6)
from
  dual;
SUBSTR
106071It looks like you want to move two positions to the right rather than a single position:
select
  substr('F15047/proc_10607161001.csv', instr('F15047/proc_10607161001.csv', '_', 1, 1) + 2, 6)
from
  dual;
SUBSTR
060716Hope that helps a bit,
Mark