Using date to get tomorrows date in Bash - bash

I want to write a bash script that will run on a given but process data with next days date, My current approach is to get the unix time stamp and add a days worth of seconds to it, but I cant get it working, and haven't yet found what I'm looking for online.
Here's what I've tried, I feel like the problem is that its a string an not a number, but I dont know enough about bash to be sure, is this correct? and how do I resolve this?
today="$(date +'%s')"
tomorrow="$($today + 86400)"
echo "$today"
echo "$tomorrow"

If you have gnu-date then to get next day you can just do:
date -d '+1 day'

Some of the answers for this question depend on having GNU date installed. If you don't have GNU date, you can use the built-in date command with the -v option.
The command
$ date -v+1d
returns tomorrow's date.
You can use it with all the standard date formatting options, so
$ date -v+1d +%Y-%m-%d
returns tomorrow's date in the format YYYY-MM-DD.

$(...) is command substitution. You're trying to run $today + 86400 as a command.
$((...)) is arithmetic expansion. This is what you want to use.
tomorrow=$(( today + 86400 ))
Also see http://mywiki.wooledge.org/ArithmeticExpression for more on doing arithmetics in the shell.

I hope that this will solve your problem here.
date --date 'next day'

Set your timezone, then run date.
E.g.
TZ=UTC-24 date
Alternatively, I'd use perl:
perl -e 'print localtime(time+84600)."\n"'

echo $(date --date="next day" +%Y%m%d)
This will output
20170623

I like
input_date=$(date '+%Y-%m-%d')
tomorrow_date=$(date '+%Y-%m-%d' -d "$input_date + 1 day")
echo "$tomorrow_date"
It returns the date of the day after $input_date, in format YYYY-MM-DD

You can try below
#!/bin/bash
today=`date`
tomorrow=`date --date="next day"`
echo "$today"
echo "$tomorrow"

Related

Script shell check validate date format YYYYMMDDHHMMSS

i want to check if Date is valid or not using Script shell.
I tried this:
date "+%d/%m/%Y" -d "09/10/2013" > /dev/null 2>&1
is_valid=$?
It works fine for me when it comes to dd/MM/YYYY format. I Tried
date "+%Y%m%d%H%M%S" -d "20191001041253" > /dev/null 2>&1
is_valid=$?
for YYYYmmddHHMMSS but it doesn't work.
Could someone please guide me to check if a date is valid or not with the YYYYmmddHHMMSS.
GNU date has a very relaxed syntax as to how to read input. The "+..." is only for specifying the output of it, you cannot in any way specify how should it read/parse the input.
You need to first convert your input into something GNU date can understand. I found that the format mentioned in the DATE STRING section in man page seems to be most reliable.
s=20191001041253
# 2019-10-01 04:12:53
s="${s::4}-${s:4:2}-${s:6:2} ${s:8:2}:${s:10:2}:${s:12:2}"
if date --date="$s" >/dev/null 2>/dev/null; then
echo "Its ok"
else
echo "its invalid"
fi
Note that bsd date has an -f options that allows specifying input format. I wish such option would be available with gnu date.
The 'date' utility accept date values in many formats (look at info date, 'Date Input formats' section). The bad news is that the format you need ( 'YYYYmmddHHMMSS') is not supported. The good news is that similar format (ISO compact format 'YYYYmmdd< HH:MM:SS') is supported.
date -d '20191001 05:06:07'
Tue Oct 1 05:06:07 IDT 2019
If you want to LIMIT the input date format to accept ONLY YYYYmmddHHMMSS, you actually have to write a small filter
input=20191001050607
# Convert to YYYYmmdd HH:MM:SS
t="${input:0:8} ${input:8:2}:${input:10:2}:${input:12:2}"
is_valid=
# Validate; Check for 14 digits + try to convert to date
[[ "$input" =~ ^[0-9]{14}$ ]] && date -d "$t" > /dev/null 2>&1 && is_valid=YES
echo "$is_valid"
Try changing it to ISO format:-
$ date "+%Y%m%d%H%M%S" -d "2019-10-01 04:12:53"
20191001041253
$ echo $?
0

How to subtract 60 minutes from current time in unix

I'm currently creating a shell script that will run a python code once an hour that collects, processes, and displays data from a radar for the previous hour.
The python code I am using requires a UTC begin time and end time in format "YYYYMMDDHHmm". So far, I have found using the unix command date -u +"%Y%m%d%H%M" will retrieve my current time in the correct format, but I have not been able to find a way to subtract 60 minutes from this first time and have it output the "start" time/
code I have tried:
date +"%Y%m%d%H%M-60" >> out: 201908201833-60
now= date -u +"%Y%m%d%H%M" >> out:201908201834
echo "$now - 60" >> out: - 60
I'm just starting to self teach/learn shell coding and I am more comfortable with python coding which is why my attempts are set up more like how you would write with python. I'm sure there is a way to store the variable and have it subtract 60 from the end time, but I have not been able to find a good online source for this (both on here and via Google).
You can use -d option in date:
date -u +"%Y%m%d%H%M" -d '-60 minutes'
or else subtract 1 hour instead of 60 minutes:
date -u +"%Y%m%d%H%M" -d '-1 hour'
To be able to capture this value in a variable, use command substitution:
now=$(date -u +"%Y%m%d%H%M" -d '-1 hour')
On OSX (BSD) use this date command as -d is not supported:
now=$(date -u -v-1H +"%Y%m%d%H%M")
Your current attempt has some simple shell script errors.
now= date -u +"%Y%m%d%H%M" >> out:201908201834
This assigns an empty string to the variable now and then runs the date command as previously. If the plan is to capture the output to the variable now, the syntax for that is
now=$(date -u +"%Y%m%d%H%M")
Next up, you try to
echo "$now - 60"
which of course will output the literal string
201908201834 - 60
rather than perform arithmetic evaluation. You can say
echo "$((now - 60))"
to subtract 60 from the value and echo that -- but of course, date arithmetic isn't that simple; subtracting 60 from 201908210012 will not produce 201908202312 like you would hope.
If you have GNU date (that's a big if if you really want to target any Unix) you could simply have done
date -u -d "60 minutes ago" +%F%H%M
but if you are doing this from Python anyway, performing the date extraction and manipulation in Python too will be a lot more efficient as well as more portable.
from datetime import datetime, timedelta
dt = datetime.strptime(when,'%Y%m%d%H%M')
print(dt - timedelta(minutes=60))
The shell command substitution $(command) and arithmetic evaluation $((expression)) syntaxes look vaguely similar, but are really unrelated. Both of them have been introduced after the fundamental shell syntax was already stable, so they had to find a way to introduce new syntax which didn't already have a well-established meaning in the original Bourne shell.

Bash -1 year from date

Year=`date '+%Y'`
RTRN1=$?
This returns the current date in the logs, however i want to return the year before, so instead of this returning 2017 i want 2016.
Any help appreciated! Thanks
For GNU date utility: use -d (--date) option to adjust the date:
Year=$(date +%Y -d'1 year ago')
echo $Year
2016
As we have bash, it's possible to use let
let YEAR=`date +%Y`-1
echo $YEAR
You can always capture the year with date +'%Y'. You can subtract 1 with the POSIX arithmetic operator, e.g.
$ echo $(($(date +%Y) - 1))
2016
You can also use the POSIX compliant expr math operators, e.g.
$ expr $(date +%Y) - 1
2016
(note: with expr you must leave a space between the math operator and the values)
The GNU date operator -d with '1 year ago' will work as specified in the comments and other answer, along with let dt=$(date +%Y)-1; echo $dt as specified in the other answer (no spaces allowed with let).
Of all the choices, if I didn't have GNU date, I'd pick the POSIX arithmetic operator $((...)) with a date command substitution minus 1.

Getting date of X days ago in bash script, using argument variable

I'm trying to calculate the date for a dynamic number of days ago in a bash script.
This is what I've done -
#!/bin/bash
STAMP=`date --date='$1 day ago' +%y%m%d`
but when running myscript 2, it says -
date: invalid date `$1 day ago'
How can I use my argument value in this formula?
It works if ' is replaced with " into this command on the script -
STAMP=`date --date="$1 day ago" +%y%m%d`
The clue was the two different character ` and ' used in the error response -
date: invalid date `$1 day ago'
An expert in bash scripting (not me) can probably explain why this has happen.
It's because variable substitution wouldn't happen in single quotes, i.e. '$1' wouldn't expand but "$1" would.
As such, saying
STAMP=`date --date="$1 day ago" +%y%m%d`
or
STAMP=$(date --date="$1 day ago" +%y%m%d)
would work.

Shell script datetime function?

I am running the following script:
#!/bin/ksh
./clear_old
./rooms_xls.pl 2/23/09
cd doors
./doors_xls.pl 2/23/09
How can I get the date to be today's date based upon the system/server time?
This should work fine (and be almost compliant with your format):
#!/bin/ksh
./clear_old
./rooms_xls.pl `date +%D`
cd doors
./doors_xls.pl `date +%D`
$(date)
works
Use backticks with the date command:
./rooms_xls.pl `date +%m/%d/%Y`

Resources