Subtract 1 hour from date in UNIX shell script
ShellDatetimeUnixShell Problem Overview
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"`
Shell Solutions
Solution 1 - Shell
The following command works on recent versions of GNU date:
date -d '1 hour ago' "+%m/%d/%Y -%H:%M:%S"
Solution 2 - Shell
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 $(( epoch - 60*60 )) is bash-aritmetics - subtracting 1hour in seconds. Prints:
current date: 04/20/2017 -18:14:31
date - 60min: 04/20/2017 -17:14:31
Solution 3 - Shell
if you need substract with timestamp :
timestamp=$(date +%s -d '1 hour ago');
Solution 4 - Shell
This work on my Ubuntu 16.04 date:
date --date="@$(($(date +%s) - 3600))" "+%m/%d/%Y -%H:%M:%S"
And the date version is date (GNU coreutils) 8.25
Solution 5 - Shell
$ date +%Y-%m-%d-%H
2019-04-09-20
$ date -v-1H +%Y-%m-%d-%H
2019-04-09-19
But in shell use as like date +%Y-%m-%d-%H, date -v-1H +%Y-%m-%d-%H
Solution 6 - Shell
Convert to timestamp (a long integer), subtract the right number of milliseconds, reformat to the format you need.
Hard to give more details since you don't specify a programming language...
Solution 7 - Shell
If you need change timezone before subtraction with new format too:
$(TZ=US/Eastern date -d '1 hour ago' '+%Y-%m-%d %H:%M')
Solution 8 - Shell
Here another way to subtract 1 hour.
yesterdayDate=`date -d '2018-11-24 00:09 -1 hour' +'%Y-%m-%d %H:%M'`
echo $yesterdayDate
Output:
2018-11-23 23:09
I hope that it can help someone.
Solution 9 - Shell
DATE=date -1H "+%m/%d/%Y -%H:%M:%S"