hi,
i have 2 dates in the form: '20080315120030' and '20080310140030'. i.e. YYYYMMDDHHMMSS.
i need a way of getting the difference between them using shell script.
any thoughts?
note i only have the following date utility:
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
at present i have this:
for file in `find ${my_dir}/ -name "*_statsfile_??????????????.gz" -print`
do
ONEDAY=86400
NOW=$(date -u +%Y%m%d%H%M%S)
FILEDATE=\`echo $file | sed -e 's/^.*\_stats_//g' -e 's/\\.gz$//g'\`
\(\(DIFF=$NOW - $FILEDATE\)\)
CALC_DIFF=$\(\($DIFF / $ONEDAY\)\)
echo "[$file], Current Time: $NOW. File Time: $FILEDATE. Difference: $CALC_DIFF."
done
both dates need to be in UTC. Is there a way to make them BOTH UTC. the date 'NOW' is UTC but the 'FILEDATE' isnt. 
It's possible but it would help if you could say which time zone the second one is.
The standard solution would be to convert them both to a canonical format (maybe epoch, that is, seconds since Jan 1 1970) and then the problem is just how to format the result back as a human-readable time expression. In what form would you like the output?
I would use Perl or Python because they come with handy libraries which can do the grunt work for you.
at the moment the first date, NOW is using date -u (UTC) and the file names will also be UTC. At present i have created dummy files with made up time stamps.
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?
im only after the number of days between the 2 dates.
If you have control over the file names then simply compare the date stamp in the file's name with the output of
date -d "3 days ago" -u +%Y%m%d%H%M%S
thats where the problem is, i get the following error:
date: illegal option -- d
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
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 :rolleyes:
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
Actually the "Similar threads" list at the bottom of this page has some nice links. Weird date difference problem looks like d�j� vu all over again.
This is the Linux forum, otherwise I would not even have suggested date -d. /-:
Why don't use find to do the job?
find <dir> -name <files> -mtime <+n> -exec rm -f {} \;
Check the manpage for more details of the used options.
Regards
The OP is not on a GNU based platofrm I guess. try perl:
#!/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. 
Hmm. Maybe the coordinates I had were correct.... or not -
here is an updated version
#!/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
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)
Using Jim's clever way of working out the seconds i ended up getting it to work using Perl's Time::Local library.
Here is my eventual code:
#!/bin/ksh
epoch_sec()
{
perl -e '
use Time::Local;
$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);
$time = timelocal($sec,$min,$hour,$mday,$mon,$year);
print int $time;
' "$1"
}
time_now=$( date -u "+%Y%m%d%H%M%S" )
echo ""
echo "Deletion Progress. Directory ${my_dir}. Time Now: $time_now."
echo "========================================================================================"
file_count=0
delete_file_count=0
for file in `find ${my_dir}/ -name "*_report_??????????????.gz" -print`
do
file_count=`expr $file_count + 1`
filedate=`echo $file | sed -e 's/^.*_report_//g' -e 's/\.gz$//g'`
today=$( date -u "+%Y%m%d%H%M%S" )
# Calculate Epochs
timenow_epoch=$( epoch_sec $today )
file_epoch=$( epoch_sec $filedate )
# Work out difference
diff=$(( $timenow_epoch - $file_epoch ))
# 259200 = 3 days
if [[ $diff -gt 259200 ]] ; then
echo ""
echo "File: [$file]"
echo " File Date: $filedate"
echo " Remove [$file]!."
echo ""
delete_file_count=`expr $delete_file_count + 1`
rm -f $file
fi
done
I didnt use -mtime Franklin52 as the file names contain the dates which need to be used NOT the modification date :).
Once again many thanks to Jim and Era and others who helped me 
