How to replace sqlldr values - oracle

Question

My raw file like below
bbbb 1000 Open 1000 Dep 12/03/2010
dddd 1001 Open 2000 Loan 13/01/2019
eeee 1003 Dor 3000 Dep 11/04/1965
Sqlldr control file
load data
infile *
truncate into table Mytab
fields
trailing nullcols
(
Name position (1:4),
Acc position(6:9),
Status Position(11:14),
Amt position(15:18),
type position(20:23),
Date position(24:33)
)
begin data
I need date column with hyphen(-) symbol not for slash(/) symbol.
12/03/2010 -- 12-03-2010
Date position(24:33) ---> What I have to mention here

Answer 1

It would help if you also provided mytab description. Because, "Date" column (whose name is invalid; it is reserved for datatype name. I renamed it to "Datum") should be of date datatype, while it seems you use varchar2. In either case, you'd use appropriate function, enclosed into double quotes. I used replace as it does what you asked for.
Here's an example:
Control file (yours is invalid as it doesn't follow data format; I fixed it):
load data
infile *
truncate into table Mytab
fields
trailing nullcols
(Name position(1:4),
Acc position(6:9),
Status position(11:14),
Amt position(16:19),
type position(21:24),
Datum position(26:35) "replace(:datum, '/', '-')"
)
begindata
bbbb 1000 Open 1000 Dep 12/03/2010
dddd 1001 Open 2000 Loan 13/01/2019
eeee 1003 Dor 3000 Dep 11/04/1965
Loading session & result:
SQL> $sqlldr scott/tiger#xe control=test34.ctl log=test34.log
SQL*Loader: Release 11.2.0.1.0 - Production on Sri O×u 24 07:54:46 2021
Copyright (c) 1982, 2009, Oracle and/or its affiliates. All rights reserved.
Commit point reached - logical record count 3
SQL> select * from mytab;
NAME ACC STATUS AMT TYPE DATUM
---------- ---------- ---------- ---------- ---------- ----------
bbbb 1000 Open 1000 Dep 12-03-2010
dddd 1001 Open 2000 Loan 13-01-2019
eeee 1003 Dor 3000 Dep 11-04-1965
SQL>