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date formatting in perl

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# 1  
Old 11-02-2011
date formatting in perl
my code:
Code:
$dateformat = yyyy_mm_dd
if($dateformat =~ m/yyyy/i)
{
print ("found");
$yyyy = strftime "%Y", localtime;
print ("year is $yyyy\n");
$dateformat =~ s/yyyy/$yyyy/;
print ("new date format is $dateformat\n");
}
elsif($dateformat =~ m/yy/i)
{
print ("found");
$yy = strftime "%y", localtime;
print ("year is $yy\n");
$dateformat =~ s/yy/$yy/;
print ("new date format is $dateformat\n");
}

output = 2011_mm_dd
Is there any method to do this in a simpler way or using sed? Please suggest
Last edited by pludi; 11-02-2011 at 06:45 AM..
# 2  
Old 11-02-2011
Code:
$dateformat = yyyy_mm_dd
if($dateformat =~ m/yyyy/i)
{
print ("found");
$yyyy = strftime "%Y", localtime;
print ("year is $yyyy\n");
$dateformat =~ s/yyyy/$yyyy/;
print ("new date format is $dateformat\n");
}
elsif($dateformat =~ m/yy/i)
{
print ("found");
$yy = strftime "%y", localtime;
print ("year is $yy\n");
$dateformat =~ s/yy/$yy/;
print ("new date format is $dateformat\n");
}

you are replacing the year only. you are replacing the month and date
# 3  
Old 11-02-2011
replacing only the year.
eg: It ll either be "2011" or "11"
# 4  
Old 11-02-2011
Not much simpler, just saves some duplicated code.

Code:
$dateformat = 'yyyy_mm_dd';
$fmt = ($dateformat =~ m/yyyy/i) ? '%Y' : $dateformat =~ m/yy/i ? '%y' : '';

if ($fmt ne '')
{
    print ("found");
    $year = strftime $fmt, localtime;
    print ("year is $year\n");
    $dateformat =~ s/$1/$year/;
    print ("new date format is $dateformat\n");
}

# 5  
Old 11-03-2011
Thanks this works perfect Smilie
can you explain the below syntax and how this works?
Code:
$fmt = ($dateformat =~ m/yyyy/i) ? '%Y' : $dateformat =~ m/yy/i ? '%y' : '';

Last edited by Franklin52; 11-03-2011 at 04:05 AM.. Reason: Please use code tags, thank you
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