Help on regex

[rajiv07]

The way you are doing it is OK, but are you sure you need to check the year, month, and day?

Thanks Kevin

Now i have try some other method.I am using Time::Local module to find the record between the date range.Is it better way to sort this problem.

[CODE=perl]#!/usr/bin/perl
use Time::Local;

my $cUser_Name="ra jiv101";
my @file;

my $cYear1=2008;

my $cYear2=2008;

my $cMon1=2;

my $cMon2=2;

$cMon1-=1;

$cMon2-=1;

my $cDay2=18;

my $cDay1=18;

my $s=0;
my $m=0;
my $h=0;

while(<DATA>)
{

next if /^(\s)*$/;

my $date=(split(/#/,$_,3))[1];

local ($year,$mon,$da y)=split(/\-/,$date);

$mon-=1;

if((timelocal($ s,$m,$h,$day,$m on,$year) >= timelocal($s,$m ,$h,$cDay1,$cMo n1,$cYear1))&&( timelocal($s,$m ,$h,$day,$mon,$ year) <= timelocal($s,$m ,$h,$cDay2,$cMo n2,$cYear2)))
{
push(@file,$_) ;
}

}

print @file;





__END__
rajiv101#2008-02-16#10:06:00#19: 49:33#127.0.0.1 #19:49:54#18:18 :00#1 27.0.0.1#CY#Y

rajiv101#2008-02-17#10:06:00#19: 49:33#127.0.0.1 #19:49:54#18:18 :00#1 27.0.0.1#CY#Y

rajiv101#2008-02-17#10:06:00#19: 49:33#127.0.0.1 #19:49:54#18:18 :00#1 27.0.0.1#CY#Y

rajiv101#2008-02-18#10:06:00#19: 49:33#127.0.0.1 #19:49:54#18:18 :00#1 27.0.0.1#CY#Y

rajiv101#2008-02-18#10:06:00#19: 49:33#127.0.0.1 #19:49:54#18:18 :00#1 27.0.0.1#CY#Y

rajiv101#2008-02-19#10:06:00#19: 49:33#127.0.0.1 #19:49:54#18:18 :00#1 27.0.0.1#CY#Y[/CODE]

Regards
Rajiv

[KevinADC]

It probably is better, but since I have no idea what you are doing with your script besides searching a range of dates it is hard to say if it is the best way to do it. Make sure to read the Time::Local documentation thoroughly as it looks like you might be making an error in calculating the month. The months are number: 0 to 11 instead of 1 to 12.