I tried a lot to load data to table from excel(.csv) using sql*loader the oracle version of sql*loader doesn't support the control file created using notepad(.ctl).Though i given a filename with extension as .ctl it seems as a .txt file. Is there any alternate way to create it?
View 3 RepliesI am not able to create control file from scrip . it showing following error
ORA-01081: cannot start already-running ORACLE - shut it down first
CHARACTER SET WE8ISO8859P1
*
ERROR at line 19:
ORA-02236: invalid file name
how to create control file and how to load the data through command window in our database using sql * loader.i am having structure in my database and .csv file in my desktop.
View 20 Replies View RelatedIs it possible to use where clause to discard one of the item which is not able to fit into column because of the length constraint.
Example:
1.Remove first digit from the item_ID where ITEM_ID IN (12345)
2.Do not load data WHERE ITEM_ID IN (12345)
I'm having a problem with SQL loader and the control file. I want to load a delimited file. The script will eventually be automated where the file name is passed in to the script, it's not a static name.
It's a simple SQL loader Unix script that I have created as follows
Unix file called test_load
# Auto Load
#
#
export data_file=/dev/test_$1$2.csv;
export ORACLE_HOME=/u01/oracle/product/10.2.0;
export ORACLE_SID=XXX;
export PATH=$PATH:$ORACLE_HOME/bin;
sqlldr userid=XXX/XXX data=$data_file
control=/dev/cntrl/test.ctl
errors=99999
bad=/dev/bad/test_$1$2_$$.bad
log=/dev/logs/test_$1$2_$$.log
the top of my control file is as follows
load data
truncate
into table test
fields terminated by "|"
when record_type = 'AA'
(
running at the prompt ./test_load myload 20100319
The following error occurs
SQL*Loader-350: Syntax error at line 5.
Expecting "(", found keyword when.
when record_type = 'AA'
^
I believe the format of my control file is correct but for some reason it won't load.
I am trying to create a new controlfile. I used "alter database backup controlfile to trace;" and then I copied and pasted the contents of the trace file into the .sql file. My sql script is
CREATE CONTROLFILE SET DATABASE "FIVESTAR" RESETLOGS NOARCHIVELOG
MAXLOGFILES 16
MAXLOGMEMBERS 3
MAXDATAFILES 100
MAXINSTANCES 8
MAXLOGHISTORY 292
[code]....
Actually, the name of my database was 'STAR' and I changed it to "FIVESTAR" int this .sql script. Now when I run this script with SQL> @ <path of my sql script>, it gives me the following errors:
ERROR at line 1:
ORA-01503: CREATE CONTROLFILE failed
ORA-01504: database name 'FIVESTAR' does not match parameter db_name 'STAR'
I have a csv file extracted from mainframe which has to be loaded into oracle using sqlldr utility.The numbers are in the format +0000003333, -0000003232.44 etc
I have to convert it to 3333 and -3232.44 and insert into the table.
I have used syntax like
Load file....append into table (t_num expression "to_number(':tnum,'99999.999')")
This gives me an invalid number error.
The reason "why sql loader's CONTROL FILE is not working, if that location has space in between"
E.g. - C:practice for sqlloadercontrol1.ctl
But it works if we remove the space in between the path as below:
E.g. - C:practice_for_sqlloadercontrol1.ctl
I am trying to have sqlldr running against a file:
C:oratest20080318
Is it possible I get SQL*Loader-500: Unable to open file and 503 file not found just because the file name does not have an extension?
I can see that any file name I try it looks after whateverFileName.dat! Is there a way to have sqlldr working with files that do not have extensions?
I am loading data using sqlldr command in UNIX to an oracle table and want to concatenate timestamp to a file name in the "create_file_name" column in the code below.
I have the below code within the control file..
LOAD DATA
TRUNCATE
INTO TABLE TABLEA
TRAILING NULLCOLS
(
file_type POSITION(1:5) CHAR,
business_date POSITION(16:23) DATE "YYYYMMDD",
create_file_name "FILE_NAME" EXPRESSION "SELECT TO_CHAR(CURRENT_TIMESTAMP(3), 'YYYYMMDDHH24MISS') FROM DUAL")
The load fails with SQL Loader error: "Expecting valid column specification, ",", ")", found keyword EXPRESSION found instead of column. How the timestamp to a filename can be appended?
I am using perl script to dynamically generate the control file.If I have data in the control file as well as in the datafile, how would i write the control file in that case. Is the below one correct?
load data
INFILE '*'
INFILE '/export/home/test/test.csv'
INSERT INTO TABLE EMP fields
terminated by "," optionally enclosed by '"'
trailing nullcols
( empno, empname, sal, deptno )
[code]....
Is there any way that if my control file contains half of the data and my data file contains the other half of the data can i club this data into a logical record in the control file to populate the DB?
My exact 2nd requirement is, my DB contains 5 cols and for 1 col the data is common(countryName) which i have to pass to the control file dynamically and the .csv file contains the data for the other four cols. How could i combine these in the ctrl file and populate the DB?
so if the DB contains CountryName, empid, ename, sal and dept..I will get the CountryName to the ctrl file and csv contains the data for empid, ename, sal and dept. How would i combine these data into a logical record and populate the DB?
I am trying to generate dynamic control file, as the files I want to upload are coming from different source and their name is constantly changing but following a fix pattern and naming convention.
I am able to generate dynamic control file through SQL. But while calling from BATCH file, i am unable to sent the file name as parameter.
All the examples i have searched are for UNIX, how to do it with BATCH File in WINDOWS.
I have a control file like following:
LOAD DATA
INFILE *
INTO TABLE member
REPLACE
FIELDS TERMINATED BY '|' OPTIONALLY ENCLOSED BY '"'
(
control_id,
name,
address
)
problem is that control_id value is not in data file and I have to assign to each row the same value generated from sequence or from unix variable.
For example, after I run sqlldr, I have to have records in the table like following:
control_id name address
---------- ---- -------
1847 Charlie 250 yonge st
1847 Peter 5 Brookbanks dr
1847 Ben 123 King st
.
.
.
How do I do that?
My oracle table having 2 fields.
filed1 VARCHAR(500)
field2 NUMBER.
i load data to this table from a file using sqlldr.
what is the proper data type should i use in control-file for both the fields.? i dont mention any datatye in ctl file which is working fine with given dataset.
I have exported and imported a schema from one server to another. In the source schema, I have a public synonym. I do not know the name of that synonym. In the destination schema, the public synonym is missing. How to create the public synonym which is missing in the destination database?In the source, I queried dba_synonyms, all_synonyms... but it returns no rows selected.
View 3 Replies View RelatedThe following control file updates multiple rows in database table.
LOAD DATA
INFILE *
replace
INTO TABLE temp_tab
FIELDS TERMINATED BY ","
(Data LOBFILE(CONSTANT cadd_pass.xml) terminated by eof
There are 21 lines in the xml. So 21 rows are updated in table.update only one row?
My Table structure
column1 varchar(10)
column2 Date
Column3 varcahr(2)
Column4 varcahr(2)
By Data file
asds 12/12/2001asas
textsd asds
asds 12/12/2001asas
ramkiy asds
I still want to insert row 2 and row 4 into table by defaulting the date. how can I handle this in control file?
I have a table revenue
create table revenue
(
person varchar2(23),
month varchar2(3),
rev_amt number
)
and i have data in a file like below
Person Jan Feb Mar Apr Mai Jun Jul Aug Sep Oct Nov Dez
--------------------------------------------------------
Schnyder,345,223,122,345,324,244,123,123,345,121,345,197
Weber,234,234,123,457,456,287,234,123,678,656,341,567
Keller,596,276,347,134,743,545,216,456,124,753,346,456
Meyer,987,345,645,567,834,567,789,234,678,973,456,125
Holzer,509,154,876,347,146,788,174,986,568,246,324,987
Müller,456,125,678,235,878,237,567,237,788,237,324,778
Binggeli,487,347,458,347,235,864,689,235,764,964,624,347
Stoller,596,237,976,876,346,567,126,879,125,568,124,753
Marty,094,234,235,763,054,567,237,457,325,753,577,346
Studer,784,567,235,753,124,575,864,235,753,864,634,678
i want to load it into the table in the following way.
Person Month Revenue
-------------------------
Schnyder Jan 345
Schnyder Feb 223
Schnyder Mar 122
Schnyder Apr 345
Schnyder Mai 324
Schnyder Jun 244
Schnyder Jul 123
Schnyder Aug 123
Schnyder Sep 345
Schnyder Oct 121
Schnyder Nov 345
Schnyder Dez 197
........ ... ...
How to write control file to load this data into the above revenue table.
I'm unable to do an import of a *.dmp file.
[oracle@oracledbserver ASG1]$ cd /media/volume-01/u01/app/oracle/product/
[oracle@oracledbserver product]$ ls
11.2.0 20-04-2013full_backup.dmp full01-03-2013_backup.dmp new.dmp today.dmp
[oracle@oracledbserver product]$
[oracle@oracledbserver product]$
[oracle@oracledbserver product]$
[oracle@oracledbserver product]$ impdp full=Y directory=agge_dir dumpfile=/media/volume-01/u01/app/oracle/product/new.dmp NOLOGFILE=y;
Import: Release 11.2.0.1.0 - Production on Tue May 7 16:51:47 2013
Copyright (c) 1982, 2009, Oracle and/or its affiliates. All rights reserved.
Username: sys as sysdba
Password:
Connected to: Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production
With the Partitioning, OLAP, Data Mining and Real Application Testing options
ORA-39001: invalid argument value
ORA-39000: bad dump file specification
ORA-39088: file name cannot contain a path specification
[oracle@oracledbserver product]$
This is Oracle 11g hosted on an eucalyptus cloud instance.
I am recieving errors when trying to load the control file. The errors are as follows:
SQL*Loader-500 Unable to open file (homework.ctl)
SQL*Loader-553 file not found
SQL*Loader-559 SYstem error: The system cannot find the file specified.
My control file is located directly in the C drive (C:homework.ctl). The control file contains the following
LOAD DATA
INFILE 'c:country.dat'
APPEND INTO TABLE homework
fields terminated by ',' optionally encloded by '"'
(country, month, day)
WHEN (month='April')
The command I am entering is:
sqlldr system/password control=homework.ctl
I've tried c:homework.ctl, 'c:homework.ctl', and placing the file in the BIN folder of Oracle.
while importing data i got this error in my log file.and i cannot import my data successfuly
in my log file error i found like this
ORA-01658: unable to create INITIAL extent for segment in tablespace USERS
IMP-00017: following statement failed with ORACLE error 1658:
IMP-00003: ORACLE error 1658 encountered
ORA-01658: unable to create INITIAL extent for segment in tablespace USERS
IMP-00017: following statement failed with ORACLE error 1658:
i can import my data using imp utility using this syntax
C:UsersAdministrator>imp tiger/****@tcs file=E:DUMP s.
dmp log=E:DUMP s.log fromuser=tiger121 touser=tiger statistics=none
this my user tiger default tablespace its uses and its a auto extend on and locally managed tablespace,and i have enough space on my drive also.
Are there any GUI based tools that can auto generate a CTL file based off a CSV input? I'd love something like this since I have quite a few SQL*LDR projects coming up!
View 2 Replies View RelatedI'm getting an error when trying to use the new Data Pump Export/Import utility.
I am able to create a directory using SQLPLus, and I get the "Directory Created" message, but no directory actually gets created on the server.
SQL> CREATE DIRECTORY datapump AS 'C:Inetpubdatafiledatapump';
Directory created. But I dont see the directory created on the server.
Then on the server:
C:Documents and SettingsAdministrator>expdp ******/****** FULL=y DIRECTORY=datapump DUMPFILE=expdata.dmp LOGFILE=expdata.log
Export: Release 10.2.0.1.0 - Production on Wednesday, 01 November, 2006 1:51:55
Copyright (c) 2003, 2005, Oracle. All rights reserved.
Connected to: Oracle Database 10g Enterprise Edition Release 10.2.0.1.0 - Production
With the Partitioning, OLAP and Data Mining options
ORA-39002: invalid operation
ORA-39070: Unable to open the log file.
ORA-29283: invalid file operation
ORA-06512: at "SYS.UTL_FILE", line 475
ORA-29283: invalid file operation
By mistakenly Oradata file has been deleted and there is I have not any backup when we Startup the database then we get an Error as
ORA-00205: error in identifying control file, check alert log for more info
Now I want to create controlfile as
Microsoft Windows [Version 6.1.7600]
Copyright © 2009 Microsoft Corporation. All rights reserved.
C:\Users\SARASWATIMAA>SQLPLUS
SQL*Plus: Release 10.2.0.1.0 - Production on Sun Apr 8 23:15:09 2012
Copyright © 1982, 2005, Oracle. All rights reserved.
Enter user-name: SYS AS SYSDBA
Enter password:
[code]...
where is My Mistake in Creating Cotrolfile..I want to know that in which condition we use "SET" or "REUSE" in Create Controlfile.
I want to create binary copy of control file every time I do backup. I am using sql "alter database backup controlfile to ''+FRADG/TESTDB/controlfile/controlfile.ctl'' "; in my rman backup script. If I run it second time I got errors as it does does not allow to overwrite existing one.
View 5 Replies View RelatedWhat is difference when you issue a create controlfile with "standby" keyword on the primary database
ALTER DATABASE CREATE STANDBY CONTROLFILE AS '/path/db_stby.ctl';
vs. creating a control using
ALTER DATABASE BACKUP CONTROLFILE to '/path/db_control.ctl';
what does ORACLE put into the standby control that is "extra"? ie. what is the difference between a standby control file and a normal control file?
problem with oracle 10 g . i started service through server i got some error
ora-09925 unable to create audit trail file.
3 files are missing in admin folder these are three files adump, dpdump, pfile was missing i have not taken any backup also.
in admin file also the database folder name was missing.
i have successfully compuled and run my Developer9i form on the web but when i make a slight change in the form (add any control or do anything with form) and then recompile it it gave me error: FRM-30087: Unable to create form file why forms show this error and how it can be solved?
View 5 Replies View Related1)Is there a way to load two tables from two input files in one control file?
I have two control files.
LOAD DATA
INFILE 'C:
ame.txt'
BADFILE 'C:
ame.bad'
DISCARDFILE 'C:
[code]....
Can i load emp_tab and job_table using one control file and two input files name.txt and job.txt?
2)Is there a way to pass the path as a parameter in the control file?
In the job below, when i execute the sqlldr , can I pass C:/job.txt as an input instead of specifying it in control file?
While Running a Report through form i am getting this below error.
REP-0118 Unable to create a temporary file
But while running the same report in a windows 98 or windowsXp PC i dont get this error.