getting the date to insert into mySQL

hello
i am trying to make a log table where i insert the date into the table. how can i grab the current date in a yyyy-mm-dd format?

thanks for the help

Hi David,


you can use this following script for getting the date in yyyy-mm-dd format.

just use &date_format for passing this date format to the database.


#!/usr/bin/perl

####-----Create yyyy-mm-dd date format---####

print &date_format;


sub date_format{

($sec,$min,$hour,$wday,$month,$year,$mday)=localtime(time());

$month="0".$month if ($month)<10;
#make month in 2 digits if it is less than 2

$wday="0".$wday if ($wday)<10;
#make date in 2 digits if it is less than 2


$yr=substr($year,1);
$yr="20".$yr if length($yr)==2;

#make year in 4 digits


$date=$yr."-".$month."-".$wday;
#concatenate the date in yyyy-mm-dd format.

return $date;
}




------------------
SR -
shiju.dreamcenter.net

"The fear of the LORD is the beginning of knowledge..."

<BLOCKQUOTE><font size="1" face="Verdana,Arial,Helvetica">quote:</font><HR>Originally posted by bydavid:
hello
i am trying to make a log table where i insert the date into the table. how can i grab the current date in a yyyy-mm-dd format?
[/quote]
I'm a lazy guy so I use modules. Here is an interesting way:

use strict;
use Date::Calc qw(Today Add_Delta_Days);

print mysql_date(-1);
print mysql_date();
print mysql_date(+1);
print mysql_date(+2);

sub mysql_date {
my $offset = shift | | 0;
my @today = Today();
my @date = Add_Delta_Days(@today,$offset);
my $year = sprintf("%4d", $date[0]);
my $month = sprintf("%02d", $date[1]);
my $day = sprintf("%02d", $date[2]);
my $mysql_date = qq|$year-$month-$dayn|;
return $mysql_date;
}