how can i get the current date, convert to no. of days since Jan 1 1970. & the result is to compare against the "last changed" field in /etc/shadow to ensure the "last changed" occurs in the past. thanks for any help!

Uribo: Date arthmetic with classic unix tools (non-GNU tools) is difficult. I'm not much of a Perl programmer, but I did throw this function together that returns the number of seconds from the Epoch: #!/bin/ksh # This function takes a date time string of the format: # YYYY MM DD HH MM SS # perl autosplits the string and uses timelocal to return # the number of seconds from the Epoch. # No error checking! function seconds_from_epoch { echo $*| perl -MTime::Local -ane ' my $epochseconds = timelocal($F[5], $F[4], $F[3], $F[2], $F[1] - 1, $F[0]); print "$epochseconds\n"; ' } s1970=$(seconds_from_epoch "1970 1 1 0 0 0") echo $s1970 end=$(seconds_from_epoch "1971 1 1 0 0 0") echo $end diff=$(((end - s1970)/86400)) echo $diff # 365 days for this example # end script There's 86,400 seconds in a day. the above example is integer arithmetic only, so it doesn't return a decimal part of a day if you choose not to use midnight as the time. If you don't have Perl, there are other solutions, but they're not very nice. Regards, nails

im using bourne shell. what are the functions for converting date into string? or is there any function that will give me the difference in days between two dates? thanks!

hi again! i have the following output when i run the above script. -36000 31496400 + (end - s1970)/86400+ uribo

hi, can i do this? # Set the current month day and year. month=`date +%m` day=`date +%d` year=`date +%Y` # Subtract Jan 1 1970 from the current day. day=`expr $day - 1` month=`expr $month - 1` year=`expr $year - 1970` # If the day is 0 then determine the last # day of the previous month. if [ $day -eq 0 ]; then # Find the preivous month. month=`expr $month - 1` # If the month is 0 then it is Dec 31 of # the previous year. if [ $month -eq 0 ]; then month=12 day=31 year=`expr $year - 1` # If the month is not zero we need to find # the last day of the month. else case $month in 1|3|5|7|8|10|12) day=31;; 4|6|9|11) day=30;; 2) if [ `expr $year % 4` -eq 0 ]; then if [ `expr $year % 400` -eq 0 ]; then day=29 elif [ `expr $year % 100` -eq 0 ]; then day=28 else day=29 fi else day=28 fi ;; esac fi fi # Print the month day and year, ignoring leap year. echo $month $day $year count_month=`expr $month \* 30` count_year=`expr $year \* 365` total=`expr $count_month + $count_year + $day` echo $total exit 0 how can i refine my script to count days more accurately? the month i used 30 days & years 365. thanks!

Hi: In Bourne shell, functions are defined and integer arithmetic is performed differently from ksh. Here is my original submission modified for sh: #!/bin/sh # This function takes a date time string of the format: # YYYY MM DD HH MM SS # perl autosplits the string and uses timelocal to return # the number of seconds from the Epoch. # No error checking! seconds_from_epoch () { echo $*| perl -MTime::Local -ane ' my $epochseconds = timelocal($F[5], $F[4], $F[3], $F[2], $F[1] - 1, $F[0]); print "$epochseconds\n"; ' } s1970=`seconds_from_epoch "1970 1 1 0 0 0"` echo $s1970 end=`seconds_from_epoch "1971 1 1 0 0 0"` echo $end nosecs=`expr $end - $s1970` diff=`expr $nosecs / 86400` echo $diff # 365 days for this example Regards, nails

i must use only bourne shell. any advice on the above code? thanks.