Shell Programming and Scripting

I have got few date format patterns like "yyyymmdd", "yy_mm_dd" etc.
There can be any combination of such patterns.
I have used add_delta_days to find "yyyy", "yy", "mm", "dd" for the current date and saved them to different variables like "$y1", "$y2", "$m1" etc

In one line, i want to check if the dateformat has yyyy/yy and replace it with the corresponding variable $y1.

Here the date format has yyyy, so i want to replace it with $y1 which has current year "2011"

i need to check and replace the string with my variable
How can this be done in simple one line?

Any help will be appreciated.Thanks in advance

Something like this?

Code:

#!/bin/bash

dateformat='yymmdd';
Y1=2011;
Y2=11;
m1=11; #---- current month november
d1=05; #---- current day
dateval=$(echo $dateformat | sed "s/yyyy/$Y1/;s/yy/$Y2/;s/mm/$m1/;s/dd/$d1/")
echo $dateval

--ahamed

Thanks ahamed.
Yes but i need to do this in perl.

Oops didn't see that... Try this...

Code:

#!/usr/bin/perl
my $Y1 = 2011;
my $Y2 = 11;
my $m1 = 11; #---- current month november
my $d1 = 05; #---- current day
my $dateformat = 'yyyymmdd';
$_=$dateformat;
s/yyyy/$Y1/g; s/yy/$Y2/g; s/mm/$m1/; s/dd/$d1/;               
my $dateval=$_;
print $dateval;

--ahamed

This if fine but will make the script look big. I need to do this in a single line because there are 3 more patterns for month..."month", "mon" too.

The above four should be done in a single line.

I modified the code already... will that be fine?

--ahamed

Thanks ahamed. It works Perfect
Can you please explain me how it works?