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From: PeteOlcott <peteolc...@gmail.com>
Newsgroups: comp.databases.oracle.tools
Subject: Why is UTL_FILE.FOPEN failing?
Date: Thu, 12 Apr 2012 08:16:21 -0700 (PDT)
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INPUT:
create or replace directory filesdir as 'c:\';
grant read on directory filesdir to public;

declare
 namesfile UTL_FILE.FILE_TYPE;
begin
 --  Syntax : FOPEN ( directory alias, filename, open mode)
 namesfile  := UTL_FILE.FOPEN('FILESDIR','CASELIST.TXT','r');
end;

OUTPUT:

SQL> create or replace directory filesdir as 'c:\';

Directory created.

SQL> grant read on directory filesdir to public;

Grant succeeded.

SQL>
  1  declare
  2   namesfile UTL_FILE.FILE_TYPE;
  3  begin
  4   --  Syntax : FOPEN ( directory alias, filename, open mode)
  5   namesfile  := UTL_FILE.FOPEN('FILESDIR','CASELIST.TXT','r');
  6* end;
  7  /
declare
*
ERROR at line 1:
ORA-29283: invalid file operation
ORA-06512: at "SYS.UTL_FILE", line 488
ORA-29283: invalid file operation
ORA-06512: at line 5