subtract 1 hour from date in unix shell script

Question

I have the following in a shell script. How can I subtract one hour while retaining the formatting?

DATE=`date \"+%m/%d/%Y -%H:%M:%S\"`

Answer 1

If you need change timezone before subtraction with new format too:

$(TZ=US/Eastern date -d '1 hour ago' '+%Y-%m-%d %H:%M')

Answer 2

date -v-60M "+%m/%d/%Y -%H:%M:%S"

DATE=`date -v-60M "+%m/%d/%Y -%H:%M:%S"`

If you have bash version 4.4+ you can use bash's internal date printing and arithmetics:

printf "current date: %(%m/%d/%Y -%H:%M:%S)T\n"
printf "date - 60min: %(%m/%d/%Y -%H:%M:%S)T\n" $(( $(printf "%(%s)T") - 60 * 60 ))

The $(printf "%(%s)T") prints the epoch seconds, the $(( epoch - 60*60 )) is bash-aritmetics - subtracting 1hour in seconds. Prints:

current date: 04/20/2017 -18:14:31
date - 60min: 04/20/2017 -17:14:31