In a unix shell, how to get yesterday's date into a variable?

Question

I\'ve got a shell script which does the following to store the current day\'s date in a variable \'dt\':

date \"+%a %d/%m/%Y\" | read dt
echo ${dt}


        

Answer 1

Though all good answers, unfortunately none of them worked for me. So I had to write something old school. ( I was on a bare minimal Linux OS )

$ date -d @$( echo $(( $(date +%s)-$((60*60*24)) )) )

You can combine this with date's usual formatting. Eg.

$ date -d @$( echo $(( $(date +%s)-$((60*60*24)) )) ) +%Y-%m-%d

Explanation : Take date input in terms of epoc seconds ( the -d option ), from which you would have subtracted one day equivalent seconds. This will give the date precisely one day back.

Answer 2

dt=$(date --date yesterday "+%a %d/%m/%Y")
echo $dt

Answer 3

Try the following method:

dt=`case "$OSTYPE" in darwin*) date -v-1d "+%s"; ;; *) date -d "1 days ago" "+%s"; esac`
echo $dt

It works on both Linux and OSX.

Answer 4

Here is a ksh script to calculate the previous date of the first argument, tested on Solaris 10.

#!/bin/ksh
 sep=""
 today=$(date '+%Y%m%d')
 today=${1:-today}
 ty=`echo $today|cut -b1-4` # today year
 tm=`echo $today|cut -b5-6` # today month
 td=`echo $today|cut -b7-8` # today day
 yy=0 # yesterday year
 ym=0 # yesterday month
 yd=0 # yesterday day

 if [ td -gt 1 ];
 then
         # today is not first of month
         let yy=ty       # same year
         let ym=tm       # same month
         let yd=td-1     # previous day
 else
         # today is first of month
         if [ tm -gt 1 ];
         then
                 # today is not first of year
                 let yy=ty       # same year
                 let ym=tm-1     # previous month
                 if [ ym -eq 1 -o ym -eq 3 -o ym -eq 5 -o ym -eq 7 -o ym -eq 8 -o ym -     eq 10 -o ym -eq 12 ];
                 then
                         let yd=31
                 fi
                 if [ ym -eq 4 -o ym -eq 6 -o ym -eq 9 -o ym -eq 11 ];
                 then
                         let yd=30
                 fi
                 if [ ym -eq 2 ];
                 then
                         # shit... :)
                         if [ ty%4 -eq 0 ];
                         then
                                 if [ ty%100 -eq 0 ];
                                 then
                                         if [ ty%400 -eq 0 ];
                                         then
                                         #echo divisible by 4, by 100, by 400
                                                 leap=1 
                                         else
                                         #echo divisible by 4, by 100, not by 400
                                                 leap=0
                                         fi
                                 else
                                         #echo divisible by 4, not by 100
                                         leap=1 
                                 fi
                         else
                                 #echo not divisible by 4
                                 leap=0 # not divisible by four
                         fi
                         let yd=28+leap
                 fi
         else
                 # today is first of year
                 # yesterday was 31-12-yy
                 let yy=ty-1     # previous year
                 let ym=12
                 let yd=31
         fi
 fi
 printf "%4d${sep}%02d${sep}%02d\n" $yy $ym $yd

Tests

bin$ for date in 20110902 20110901 20110812 20110801 20110301 20100301 20080301 21000301 20000301 20000101 ; do yesterday $date; done
20110901
20110831
20110811
20110731
20110228
20100228
20080229
21000228
20000229
19991231

Answer 5

If your HP-UX installation has Tcl installed, you might find it's date arithmetic very readable (unfortunately the Tcl shell does not have a nice "-e" option like perl):

dt=$(echo 'puts [clock format [clock scan yesterday] -format "%a %d/%m/%Y"]' | tclsh)
echo "yesterday was $dt"

This will handle all the daylight savings bother.