Re: Perl date conversion

I can not reproduce your results. Are you sure your input file is what you think it is?

echo 1300483666 | perl -lne '($j1,$j2,$hour,$mday,$mon,$year,$wday) = 
+gmtime($_);$year = $year + 1900;$mon++;printf("%4s %.2d %.2d %.2d %.2
+d %6s\n",$year,$mon,$mday,$wday,$hour,1);'

2011 03 18 05 21      1
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Re^2: Perl date conversion by gio001 (Acolyte) on Mar 23, 2011 at 17:45 UTC
My message is incorrect, I want to have this output from an input of `1300483666`: `2011 03 18 05 17 1` and instead the code returns: `2011 03 18 05 21 1` Sorry for the mistake. Can you help? I suppose I can make the `gmtime($_)`be like `gmtime($_-14400)` but I thought there might be a cleaner way to resolve the date in input to my local time (EST or EDT whatever it happens to be). Can I do without the gmtime all together? Hope you can help. Thanks	[reply] [d/l] [select]
Re^3: Perl date conversion by gio001 (Acolyte) on Mar 23, 2011 at 17:58 UTC
I replaced gmtime with localtime and it seems to work properly. Thanks Any advice anyway?	[reply]
Re^4: Perl date conversion by ikegami (Patriarch) on Mar 23, 2011 at 18:34 UTC
That's correct.	[reply]