SQL & PL/SQL :: Retrieve The Path Of A Directory
Oct 27, 2011I create a directory using this:
CREATE DIRECTORY xml_data AS 'c:\mytest';
How can I retrieve the 'C:mytest' in a select statement?
I create a directory using this:
CREATE DIRECTORY xml_data AS 'c:\mytest';
How can I retrieve the 'C:mytest' in a select statement?
I started working on utl_files. When i Am working on this topic I struck with an error: Quote:ORA-29280: invalid directory path. Below are the steps I followed to work on utl_files,
STEP1: connect sys as sysdba
STEP2: @$ORACLE_HOME/rdbms/admin/utlfile.sql
STEP3: create or replace directory utl_data_files as '/home/oracle/siva/data/'; --(the directory path is already exists in the OS)
STEP4: below is the code
DECLARE
out_file UTL_FILE.file_type;
linebuf varchar2(1999);
directory_name varchar2(100):='utl_data_files';
BEGIN
out_file:=UTL_FILE.fopen(directory_name,'emp.dat','W');
for emp_cur in (SELECT * FROM scott.EMP)
[code].....
STEP5: then it's giving the below result/error info
Quote:
SQL ERRORCODE IS:-29280
SQL ERROR MSG IS:ORA-29280: invalid directory path
PL/SQL procedure successfully completed.
When i am running the below script i am getting error like Invalid directory path. how to check the path by using dbms_ouput.
error : ORA-29280: invalid directory path
ORA-06512: at line 34
DECLARE
l_out_file UTL_FILE.file_type;
g_convert_crlf BOOLEAN := TRUE;
P_DATA VARCHAR2(32767);
l_buffer VARCHAR2(32767);
l_amount BINARY_INTEGER := 32767;
l_pos INTEGER := 1;
[code].....
When i am running the below script i am getting error like Invalid directory path. how to check the path by using dbms_ouput.
error : ORA-29280: invalid directory path
ORA-06512: at line 34
DECLARE
l_out_file UTL_FILE.file_type;
g_convert_crlf BOOLEAN := TRUE;
P_DATA VARCHAR2(32767);
l_buffer VARCHAR2(32767);
l_amount BINARY_INTEGER := 32767;
[code]....
I am testing to find out that by using utl_file, if I finish reading all the entries in the file, then I encounter no_data_found, just want to make sure I get no data found when all entries in the file are read
SQL> create or replace directory my_docs as 'c:photos';
create or replace procedure myproc is
v_line VARCHAR2(2000); -- Data line read from input file
v_file UTL_FILE.FILE_TYPE; -- Data file handle
v_dir VARCHAR2(250) := 'my_docs'; -- Directory containing the data file
[Code]....
/so I execute this procedure and I get
ERROR at line 1:
ORA-29280: invalid directory path
ORA-06512: at "SYS.UTL_FILE", line 33
ORA-06512: at "SYS.UTL_FILE", line 436
ORA-06512: at "MYPROC", line 12
ORA-06512: at line 2
I want to get file directory path dynamically with out using directory in database or not hard code like below
the purpose is i need to check image in the path directory if not found i unable to generate report, is there any possible to get dynamically
function CF_URLFormula return Char is
v_handle utl_file.file_type;
v_file_dir VARCHAR2(60) := '/u002/app/applmgr/temp/';
begin
v_handle := utl_file.fopen(v_file_dir, :photo_name, 'R');
utl_file.fclose(v_handle);
RETURN v_file_dir||:photo_name;
exception
when others then
srw.message(100,SQLERRM);
return null;
end;
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(
avarchar2(10),
bvarchar2(10)
)
/
insert into test values ('A', 'B');
[Code]...
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---------
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Consider the following case ---
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=============
1 >>>>>> 0
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(
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[code]....
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[code]....
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[Code]....
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[Code]...
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