I've written a perl script that creates a file with a series of inserts
and updates, i then run that from the sql prompt.

#!/usr/bin/perl -w

$out = "/home/prod/created_update.sql";

open (FH, "/home/prod/edward.csv") or die "unable to open file";
open (FG, ">> $out") or die "unable to open file $out";

while (<FH>) {
$count++ if (/^(.*),(.*),$/i);
$line = "update TAFUSR SET BEM_VIA_EMAIL = '1' where USR = '$1';"; #
creates sql!
print FG $line,"\n\n";
}

system "chmod 777 $out";
system "chown prod:prod $out";
(-e $out) ? print "$out has been created with $count rows\n" : print
"file $out has NOT been created\n";

hope this helps

-----Original Message-----
From: Pitchai Siva (GE Infra Energy Non-GE)
[mailto:oracledba-ezmlmshield-x6879626.[Email address protected]
Sent: 31 March 2008 10:55
To: LazyDBA Discussion
Subject: is there any way to pass the UNIX variable to SQL *Loader.

Hi all,



I've created one shell script it's called the sql loader for load the
flat file into DB table. The flat file has been created by UNIX shell
script.

I wanted to pass the UNIX current directory to the sql loader (I have
assigned the current DIR into one variable) so I want to pass the
variable to sql loader. How can I do it...?







Thanks and regards,

Siva.P

Bangalore





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