When i write the following program, i am getting error like: DECLARE vInHandle UTL_FILE.file_type; vNewLine VARCHAR2(250); BEGIN vInHandle := UTL_FILE.Fopen ('D:', 'new.txt', 'R'); LOOP BEGIN UTL_FILE.get_line(vInHandle, vNewLine); dbms_output.put_line(vNewLine); END; END LOOP; UTL_FILE.fclose(vInHandle); END; Error:-DECLARE*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 line 5 Can anyone solve this.Thanks in advanceDeveloper.

Interview Candidate
Oct 11th, 2006
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Showing Answers 1 - 5 of 5 Answers

Maria Antony

Oct 14th, 2006

Hi,
If you like to access a file using UTL_FILE, you should modify your init.ora file. The file should include the directories which are allowed to be accessed while using UTL_FILE.
That is, if you like to create a file in 'D:' , you should have an entry in your init.ora file as mentioned below.
UTL_FILE_DIR='D:'
So, Just add this and try again.
Regards,
Maria Antony.

dev

Oct 17th, 2006

Hi Maria Antony,
Thanks for the reply.
I tried by giving UTL_FILE_DIR='D:' in the init.ora.71720061376 file.
but still it's having the same error.
please let me know the solution.
Thanks in advance.
Developer.

rampratap409
Profile Answers by rampratap409 Questions by rampratap409

Dec 4th, 2006

use these steps.
1 modify init.ora   add a line
        UTL_FILE_DIR='D:My_DIR'
2. login as system as sysdba
        shutdown immediate;
         startup pfile=<pfile path> like d:oracleadminorclpfileinit.ora
     oracle should startup
      create spfile from pfile;
if step 2 is not working then
alter sytem set utl_file_dir = 'd:my_dir' scope = spfile;
restart the system.
then try your programme.
create folder "my_dir" in d: drive( os level)
login system as sysdba
create direcotory my_dir as 'd:my_dir';
grant read, write on directory my_dir to scott;