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Operating Systems Linux date difference
# 8  
Old 03-17-2008
And you are on Linux? What version of date do you have (does it grok "date --version")?

Check your PATH, you might have some weird ancient BSD relic date in your PATH overshadowing the real McCoy ...? (Does "type -all date" reveal any shadows?)

If you have Perl, the Perl FAQ has a nice answer.
era
# 9  
Old 03-17-2008
Quote:
Originally Posted by era
And you are on Linux? What version of date do you have (does it grok "date --version")?

Check your PATH, you might have some weird ancient BSD relic date in your PATH overshadowing the real McCoy ...? (Does "type -all date" reveal any shadows?)

If you have Perl, the Perl FAQ has a nice answer.
my bad I'm actually on a solaris box Smilie

typing date --version threw:

date: illegal option -- -
date: illegal option -- v
date: illegal option -- e
date: illegal option -- r
date: illegal option -- s
date: illegal option -- i
date: illegal option -- o
date: illegal option -- n
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]

type -all date
date is /usr/bin/date
# 10  
Old 03-17-2008
Actually the "Similar threads" list at the bottom of this page has some nice links. https://www.unix.com/shell-programmin...e-problem.html looks like déjá vu all over again.

This is the Linux forum, otherwise I would not even have suggested date -d. /-:
era
# 11  
Old 03-17-2008
Quote:
Originally Posted by muay_tb
i basically want to delete files over 3 days old in a directory using the current date as a marker...i cant use -mtime as a person may modify the file and the modified time will not be the same as the filename. So therefore i am using the filename as the second time stamp. if you follow me?
Why don't use find to do the job?

Code:
find <dir> -name <files> -mtime <+n> -exec rm -f {} \;

Check the manpage for more details of the used options.

Regards
# 12  
Old 03-17-2008
The OP is not on a GNU based platofrm I guess. try perl:
Code:
#!/bin/ksh
sec()
{
        perl -e '
                use POSIX qw(strftime);
                $fmt = "%s";  # %s = seconds in epoch
                $mday = substr("$ARGV[0]", 7, 2);
                $mon =  substr("$ARGV[0]", 4 ,2);
                $year = substr("$ARGV[0]", 0 ,4);    
                $hour = substr("$ARGV[0]", 9 ,2);    
                $min =  substr("$ARGV[0]", 11 ,2);                 
                $sec = substr("$ARGV[0]", 13 ,2);    
                $secs =
                  strftime($fmt, $sec, $min, $hour, $mday , $mon - 1, $year - 1900, -1, -1, -1);
                print int $secs;
                ' "$1"
}

today=$( date "+%Y%m%d%H%M%S" )
today=$( sec $today )
oldat=$( sec "20080313101504" )
ddiff=$(( $today - $oldat ))
echo "$today minus $oldat = $ddiff which is \c"
if [[ $ddiff -lt 259200 ]] ; then
   echo "less than three days \c"
else
   echo "greater than three days \c"
fi  
echo " (259200 seconds)"

# 13  
Old 03-18-2008
Quote:
Originally Posted by jim mcnamara
The OP is not on a GNU based platofrm I guess. try perl:
Code:
#!/bin/ksh
sec()
{
        perl -e '
                use POSIX qw(strftime);
                $fmt = "%s";  # %s = seconds in epoch
                $mday = substr("$ARGV[0]", 7, 2);
                $mon =  substr("$ARGV[0]", 4 ,2);
                $year = substr("$ARGV[0]", 0 ,4);    
                $hour = substr("$ARGV[0]", 9 ,2);    
                $min =  substr("$ARGV[0]", 11 ,2);                 
                $sec = substr("$ARGV[0]", 13 ,2);    
                $secs =
                  strftime($fmt, $sec, $min, $hour, $mday , $mon - 1, $year - 1900, -1, -1, -1);
                print int $secs;
                ' "$1"
}

today=$( date "+%Y%m%d%H%M%S" )
today=$( sec $today )
oldat=$( sec "20080313101504" )
ddiff=$(( $today - $oldat ))
echo "$today minus $oldat = $ddiff which is \c"
if [[ $ddiff -lt 259200 ]] ; then
   echo "less than three days \c"
else
   echo "greater than three days \c"
fi  
echo " (259200 seconds)"

Thanks Jim,

I tried the above code and i noticed you had the substr co ordinates wrong.

i changed it to:

$mday = substr("$ARGV[0]", 6, 2);
$mon = substr("$ARGV[0]", 4 ,2);
$year = substr("$ARGV[0]", 0 ,4);
$hour = substr("$ARGV[0]", 8 ,2);
$min = substr("$ARGV[0]", 10 ,2);
$sec = substr("$ARGV[0]", 12 ,2);

however whenever i run the script i get:

0 minus 0 = 0 which is less than three days

the current date is: 20080318110043 and the test date is the same you used. Smilie
# 14  
Old 03-18-2008
Hmm. Maybe the coordinates I had were correct.... or not -
here is an updated version
Code:
#!/bin/ksh
sec()
{
        perl -e '
                use POSIX qw(strftime);
                $fmt = "%s";  # %s = seconds in epoch
                $mday = substr("$ARGV[0]", 6, 2);
                $mon =  substr("$ARGV[0]", 4 ,2);
                $year = substr("$ARGV[0]", 0 ,4);    
                $hour = substr("$ARGV[0]", 8 ,2);    
                $min =  substr("$ARGV[0]", 10 ,2);                 
                $sec = substr("$ARGV[0]", 12 ,2);    
                $secs =
                  strftime($fmt, $sec, $min, $hour, $mday , $mon - 1, $year - 1900, -1, -1, -1);
                print int $secs;
                ' "$1"
}
echo "today is = `date` `date "+%s"` as a sanity check"
today=$( date "+%Y%m%d%H%M%S" )
today=$( sec $today )
oldat=$( sec "20080313101504" )
ddiff=$(( $today - $oldat ))
echo "$today minus $oldat = $ddiff which is \c"
if [[ $ddiff -lt 259200 ]] ; then
   echo "less than three days \c"
else
   echo "greater than three days \c"
fi  
echo " (259200 seconds)"

output
Code:
csadev:/home/jmcnama> ddiff.pl
today is = Tue Mar 18 09:39:21 CDT 2008 1205851161 as a sanity check
1205851161 minus 1205421304 = 429857 which is greater than three days  (259200 seconds)

Last edited by jim mcnamara; 03-18-2008 at 11:39 AM..
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