utl_file.fopen and Oracle v9

Hi,

This question for Oracle version 9.
I do not use UTL_FILE_DIR = <directory name> as Oracle recommend

I use such lines:
SQL> create directory my_out as 'c:\my_dir'; Directory created.

SQL> grant read on directory my_out to public; Grant succeeded.

SQL> grant write on directory my_out to public; Grant succeeded.

After it I created such procedure :
create or replace procedure out_test
as
my_file utl_file.file_type;

f_loc varchar2(10) := 'c:\my_dir';
f_out varchar2(10) := 'out.txt';
f_mode varchar2(2) := 'w';
f_len binary_integer := 80;

f_str varchar2(20);
begin
my_file := utl_file.fopen(f_loc, f_out, f_mode, f_len);

f_str := 'Hello';
utl_file.put_line(my_file, f_str);

utl_file.fclose(my_file);
end;
/

and for line my_file := utl_file.fopen(f_loc, f_out, f_mode, f_len); I got such error:
ERROR at line 1:

ORA-29280: invalid directory path
ORA-06512: at "SYS.UTL_FILE", line 18
ORA-06512: at "SYS.UTL_FILE", line 424
ORA-06512: at "SCOTT.OUT_TEST", line 11
ORA-06512: at line 1

Can anybody propose what to do

Thanks Received on Thu Jun 19 2003 - 02:36:13 CEST