Question

I want to use REGEXP_REPLACE in PL/SQL on Oracle 10g to remove the trailing HEX-0D from a string. I found the regex from this question.

In Perl it reads as follows:

$output =~ tr/\x{d}\x{a}//d;  

or

$output =~ s/\s+\z//;

How can I translate this to PL/SQL?

Answer 1

/* PL/SQL */

declare
  in_str constant varchar2(30) := 'foo' || chr(13) || 'bar' || chr(13);
  out_str varchar2(30);
begin
  dbms_output.put_line('in_str = ' || utl_raw.cast_to_raw(in_str));

  select regexp_replace(in_str, chr(13) || '$', '') into out_str from dual;

  dbms_output.put_line('out_str = ' || utl_raw.cast_to_raw(out_str));
end;
/

/* SQL */

select utl_raw.cast_to_raw('foo' || chr(13) || 'bar' || chr(13)) as BEFORE from dual;

select utl_raw.cast_to_raw(
  regexp_replace('foo' || chr(13) || 'bar' || chr(13), chr(13) || '$', '')
) as AFTER from dual;

/* OUTPUT */

Session altered.

in_str = 666F6F0D6261720D
out_str = 666F6F0D626172

PL/SQL procedure successfully completed.

BEFORE
------------------------------------------------------------------------------
666F6F0D6261720D

AFTER
------------------------------------------------------------------------------
666F6F0D626172