UTL_FILE.FOPEN() procedure not accepting path for directory?

I am trying to write in a file stored in c:\ drive named vin1.txt and getting this error .Please suggest!

> 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-0开发者_如何学Go6512: at
> "SCOTT.SAL_STATUS", line 12 ORA-06512:
> at line 1

HERE is the code

  create or replace procedure sal_status
   (
    p_file_dir IN varchar2,
    p_filename IN varchar2)
     IS  
    v_filehandle utl_file.file_type;
    cursor emp Is
        select * from employees
        order by department_id;
    v_dep_no departments.department_id%TYPE;
     begin
         v_filehandle :=utl_file.fopen(p_file_dir,p_filename,'w');--Opening a file
         utl_file.putf(v_filehandle,'SALARY REPORT :GENERATED ON %s\n',SYSDATE);
         utl_file.new_line(v_filehandle);
         for v_emp_rec IN emp LOOP
            v_dep_no :=v_emp_rec.department_id;
            utl_file.putf(v_filehandle,'employee %s earns:s\n',v_emp_rec.last_name,v_emp_rec.salary);                    
         end loop;
        utl_file.put_line(v_filehandle,'***END OF REPORT***');
        UTL_FILE.fclose(v_filehandle);
     end sal_status;

execute sal_status('C:\','vin1.txt');--Executing

---

Since Oracle 9i there are two ways or declaring a directory for use with UTL_FILE.

The older way is to set the INIT.ORA parameter UTL_FILE_DIR. We have to restart the database for a change to take affect. The value can like any other PATH variable; it accepts wildcards. Using this approach means passing the directory path...

UTL_FILE.FOPEN('c:\temp', 'vineet.txt', 'W');

The alternative approach is to declare a directory object.

create or replace directory temp_dir as 'C:\temp'
/

grant read, write on directory temp_dir to vineet
/

Directory objects require the exact file path, and don't accept wildcards. In this approach we pass the directory object name...

UTL_FILE.FOPEN('TEMP_DIR', 'vineet.txt', 'W');

The UTL_FILE_DIR is deprecated because it is inherently insecure - all users have access to all the OS directories specified in the path, whereas read and write privileges can de granted discretely to individual users. Also, with Directory objects we can be add, remove or change directories without bouncing the database.

In either case, the oracle OS user must have read and/or write privileges on the OS directory. In case it isn't obvious, this means the directory must be visible from the database server. So we cannot use either approach to expose a directory on our local PC to a process running on a remote database server. Files must be uploaded to the database server, or a shared network drive.

---

If the oracle OS user does not have the appropriate privileges on the OS directory, or if the path specified in the database does not match to an actual path, the program will hurl this exception:

ORA-29283: invalid file operation
ORA-06512: at "SYS.UTL_FILE", line 536
ORA-29283: invalid file operation
ORA-06512: at line 7

The OERR text for this error is pretty clear:

29283 -  "invalid file operation"
*Cause:    An attempt was made to read from a file or directory that does
           not exist, or file or directory access was denied by the
           operating system.
*Action:   Verify file and directory access privileges on the file system,
           and if reading, verify that the file exists.

---

Don't forget also that the path for the file is on the actual oracle server machine and not any local development machine that might be calling your stored procedure. This is probably very obvious but something that should be remembered.

---

You need to register the directory with Oracle. fopen takes the name of a directory object, not the path. For example:

(you may need to login as SYS to execute these)

CREATE DIRECTORY MY_DIR AS 'C:\';

GRANT READ ON DIRECTORY MY_DIR TO SCOTT;

Then, you can refer to it in the call to fopen:

execute sal_status('MY_DIR','vin1.txt');

---

The directory name seems to be case sensitive. I faced the same issue but when I provided the directory name in upper case it worked.

---

You need to have your DBA modify the init.ora file, adding the directory you want to access to the 'utl_file_dir' parameter. Your database instance will then need to be stopped and restarted because init.ora is only read when the database is brought up.

You can view (but not change) this parameter by running the following query:

SELECT *
  FROM V$PARAMETER
  WHERE NAME = 'utl_file_dir'

Share and enjoy.

---

For utl_file.open(location,filename,mode) , we need to give directory name for location but not path. For Example:DATA_FILE_DIR , this is the directory name and check out the directory path for that particular directory name.