How can I print the date which is a day before current time in Bash?
DST aware solution:
Manipulating the Timezone is possible for changing the clock some hours. Due to the daylight saving time, 24 hours ago can be today or the day before yesterday.
You are sure that yesterday is 20 or 30 hours ago. Which one? Well, the most recent one that is not today.
echo -e "$(TZ=GMT+30 date +%Y-%m-%d)\n$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
The -e parameter used in the echo command is needed with bash, but will not work with ksh. In ksh you can use the same command without the -e flag.
When your script will be used in different environments, you can start the script with #!/bin/ksh or #!/bin/bash. You could also replace the \n by a newline:
echo "$(TZ=GMT+30 date +%Y-%m-%d)
$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
date +%Y:%m:%d|awk -vFS=":" -vOFS=":" '{$3=$3-1;print}'
2009:11:9
If you have BSD (OSX) date
you can do it like this:
date -j -v-1d
Wed Dec 14 15:34:14 CET 2011
Or if you want to do date calculations on an arbitrary date:
date -j -v-1d -f "%Y-%m-%d" "2011-09-01" "+%Y-%m-%d"
2011-08-31
Sorry not mentioning I on Solaris system. As such, the -date switch is not available on Solaris bash.
I find out I can get the previous date with little trick on timezone.
DATE=`TZ=MYT+16 date +%Y-%m-%d_%r`
echo $DATE
MAC OSX
For yesterday's date:
date -v-1d +%F
where 1d defines current day minus 1 day. Similarly,
date -v-1w +%F - for previous week date
date -v-1m +%F - for previous month date
IF YOU HAVE GNU DATE,
date --date="1 day ago"
More info: https://www.cyberciti.biz/tips/linux-unix-get-yesterdays-tomorrows-date.html
Use Perl instead perhaps?
perl -e 'print scalar localtime( time - 86400 ) . "\n";'
Or, use nawk and (ab)use /usr/bin/adb:
nawk 'BEGIN{printf "0t%d=Y\n", srand()-86400}' | adb
Came across this too ... insane!
/usr/bin/truss /usr/bin/date 2>&1 | nawk -F= '/^time\(\)/ {gsub(/ /,"",$2);printf "0t%d=Y\n", $2-86400}' | adb