Bash relative date (x days ago) - linux

I have a date string that I am able to parse and format with the date command from a bash script.
But how can I determine how many days ago this date was from my script? I would like to end up with a number.


Use date itself as date value for date.
Example 5 days ago:
date -d "`date`-5days"


You can do some date arithmetics:
DATE=01/02/2010
echo $(( ( $(date +%s) - $(date -d "$DATE" +%s) ) /(24 * 60 * 60 ) ))


Convert your date and now into seconds since the epoch, subtract, divide by the number of seconds in a day:
#!/bin/bash
((a = `date -d "Wed Jan 12 02:33:22 PST 2011" +%s`))
((b = `date +%s`))
echo $(( (b-a) / (60*60*24)))

Related

how to set up a cronjob to run on the last sunday of each month? [duplicate]

I need to create a CRON job that will run on the last day of every month.
I will create it using cPanel.
Any help is appreciated.
Thanks

Possibly the easiest way is to simply do three separate jobs:
55 23 30 4,6,9,11 * myjob.sh
55 23 31 1,3,5,7,8,10,12 * myjob.sh
55 23 28 2 * myjob.sh
That will run on the 28th of February though, even on leap years so, if that's a problem, you'll need to find another way.
However, it's usually both substantially easier and correct to run the job as soon as possible on the first day of each month, with something like:
0 0 1 * * myjob.sh
and modify the script to process the previous month's data.
This removes any hassles you may encounter with figuring out which day is the last of the month, and also ensures that all data for that month is available, assuming you're processing data. Running at five minutes to midnight on the last day of the month may see you missing anything that happens between then and midnight.
This is the usual way to do it anyway, for most end-of-month jobs.
If you still really want to run it on the last day of the month, one option is to simply detect if tomorrow is the first (either as part of your script, or in the crontab itself).
So, something like:
55 23 28-31 * * [[ "$(date --date=tomorrow +\%d)" == "01" ]] && myjob.sh
should be a good start, assuming you have a relatively intelligent date program.
If your date program isn't quite advanced enough to give you relative dates, you can just put together a very simple program to give you tomorrow's day of the month (you don't need the full power of date), such as:
#include <stdio.h>
#include <time.h>
int main (void) {
// Get today, somewhere around midday (no DST issues).
time_t noonish = time (0);
struct tm *localtm = localtime (&noonish);
localtm->tm_hour = 12;
// Add one day (86,400 seconds).
noonish = mktime (localtm) + 86400;
localtm = localtime (&noonish);
// Output just day of month.
printf ("%d\n", localtm->tm_mday);
return 0;
}
and then use (assuming you've called it tomdom for "tomorrow's day of month"):
55 23 28-31 * * [[ "$(tomdom)" == "1" ]] && myjob.sh
Though you may want to consider adding error checking since both time() and mktime() can return -1 if something goes wrong. The code above, for reasons of simplicity, does not take that into account.

There's a slightly shorter method that can be used similar to one of the ones above. That is:
[ $(date -d +1day +%d) -eq 1 ] && echo "last day of month"
Also, the crontab entry could be update to only check on the 28th to 31st as it's pointless running it the other days of the month. Which would give you:
0 23 28-31 * * [ $(date -d +1day +%d) -eq 1 ] && myscript.sh

What about this one, after Wikipedia?
55 23 L * * /full/path/to/command

For AWS Cloudwatch cron implementation (Scheduling Lambdas, etc..) this works:
55 23 L * ? *
Running at 11:55pm on the last day of each month.

Adapting paxdiablo's solution, I run on the 28th and 29th of February. The data from the 29th overwrites the 28th.
# min hr date month dow
55 23 31 1,3,5,7,8,10,12 * /path/monthly_copy_data.sh
55 23 30 4,6,9,11 * /path/monthly_copy_data.sh
55 23 28,29 2 * /path/monthly_copy_data.sh

You could set up a cron job to run on every day of the month, and have it run a shell script like the following. This script works out whether tomorrow's day number is less than today's (i.e. if tomorrow is a new month), and then does whatever you want.
TODAY=`date +%d`
TOMORROW=`date +%d -d "1 day"`
# See if tomorrow's day is less than today's
if [ $TOMORROW -lt $TODAY ]; then
echo "This is the last day of the month"
# Do stuff...
fi

For a safer method in a crontab based on #Indie solution (use absolute path to date + $() does not works on all crontab systems):
0 23 28-31 * * [ `/bin/date -d +1day +\%d` -eq 1 ] && myscript.sh

Some cron implementations support the "L" flag to represent the last day of the month.
If you're lucky to be using one of those implementations, it's as simple as:
0 55 23 L * ?
That will run at 11:55 pm on the last day of every month.
http://www.quartz-scheduler.org/documentation/quartz-1.x/tutorials/crontrigger

#########################################################
# Memory Aid
# environment HOME=$HOME SHELL=$SHELL LOGNAME=$LOGNAME PATH=$PATH
#########################################################
#
# string meaning
# ------ -------
# #reboot Run once, at startup.
# #yearly Run once a year, "0 0 1 1 *".
# #annually (same as #yearly)
# #monthly Run once a month, "0 0 1 * *".
# #weekly Run once a week, "0 0 * * 0".
# #daily Run once a day, "0 0 * * *".
# #midnight (same as #daily)
# #hourly Run once an hour, "0 * * * *".
#mm hh Mday Mon Dow CMD # minute, hour, month-day month DayofW CMD
#........................................Minute of the hour
#| .................................Hour in the day (0..23)
#| | .........................Day of month, 1..31 (mon,tue,wed)
#| | | .................Month (1.12) Jan, Feb.. Dec
#| | | | ........day of the week 0-6 7==0
#| | | | | |command to be executed
#V V V V V V
* * 28-31 * * [ `date -d +'1 day' +\%d` -eq 1 ] && echo "Tomorrow is the first today now is `date`" >> ~/message
1 0 1 * * rm -f ~/message
* * 28-31 * * [ `date -d +'1 day' +\%d` -eq 1 ] && echo "HOME=$HOME LOGNAME=$LOGNAME SHELL = $SHELL PATH=$PATH"

Set up a cron job to run on the first day of the month. Then change the system's clock to be one day ahead.

I found out solution (On the last day of the month) like below from this site.
0 0 0 L * ? *
CRON details:
Seconds Minutes Hours Day Of Month Month Day Of Week Year
0 0 0 L * ? *
To cross verify above expression, click here which gives output like below.
2021-12-31 Fri 00:00:00
2022-01-31 Mon 00:00:00
2022-02-28 Mon 00:00:00
2022-03-31 Thu 00:00:00
2022-04-30 Sat 00:00:00

00 23 * * * [[ $(date +'%d') -eq $(cal | awk '!/^$/{ print $NF }' | tail -1) ]] && job
Check out a related question on the unix.com forum.

You can just connect all answers in one cron line and use only date command.
Just check the difference between day of the month which is today and will be tomorrow:
0 23 * * * root [ $(expr $(date +\%d -d '1 days') - $(date +\%d) ) -le 0 ] && echo true
If these difference is below 0 it means that we change the month and there is last day of the month.

55 23 28-31 * * echo "[ $(date -d +1day +%d) -eq 1 ] && my.sh" | /bin/bash

What about this?
edit user's .bashprofile adding:
export LAST_DAY_OF_MONTH=$(cal | awk '!/^$/{ print $NF }' | tail -1)
Then add this entry to crontab:
mm hh * * 1-7 [[ $(date +'%d') -eq $LAST_DAY_OF_MONTH ]] && /absolutepath/myscript.sh

In tools like Jenkins, where usually there is no support for L nor tools similar to date, a cool trick might be setting up the timezone correctly. E.g. Pacific/Kiritimati is GMT+14:00, so if you're in Europe or in the US, this might do the trick.
TZ=Pacific/Kiritimati \n H 0 1 * *
Result: Would last have run at Saturday, April 30, 2022 10:54:53 AM GMT; would next run at Tuesday, May 31, 2022 10:54:53 AM GMT.

Use the below code to run cron on the last day of the month in PHP
$commands = '30 23 '.date('t').' '.date('n').' *';

The last day of month can be 28-31 depending on what month it is (Feb, March etc). However in either of these cases, the next day is always 1st of next month. So we can use that to make sure we run some job always on the last day of a month using the code below:
0 8 28-31 * * [ "$(date +%d -d tomorrow)" = "01" ] && /your/script.sh

Not sure of other languages but in javascript it is possible.
If you need your job to be completed before first day of month node-cron will allow you to set timezone - you have to set UTC+12:00 and if job is not too long most of the world will have results before start of their month.

If the day-of-the-month field could accept day zero that would very simply solve this problem. Eg. astronomers use day zero to express the last day of the previous month. So
00 08 00 * * /usr/local/sbin/backup
would do the job in simple and easy way.

Better way to schedule cron on every next month of 1st day
This will run the command foo at 12:00AM.
0 0 1 * * /usr/bin/foo

Be cautious with "yesterday", "today", "1day" in the 'date' program if running between midnight and 1am, because often those really mean "24 hours" which will be two days when daylight saving time change causes a 23 hour day. I use "date -d '1am -12 hour' "

Shell script to print a range of dates with a 6 hour increment

I got a shell script which prints date from one given date to another with six hour difference. I need hour 00,06,12 and 18. It does perfectly that up to some dates and then prints 01,07,13,19. I can not understand what is the reason. Here is the script:
#!/bin/bash
start=$(date --date '1 jan 1998 00:00' +%s)
stop=$(date --date '31 dec 1998 18:00' +%s)
for t in $(seq ${start} 21600 ${stop})
do
idate1=`date --date #${t} +'%Y%m%d%H'`
idate2=`date --date #${t} +'%Y%m%d'`
iyr1=`date --date #${t} +'%Y'`
imon1=`date --date #${t} +'%m'`
iday1=`date --date #${t} +'%d'`
# sleep 2s
echo $idate1
done

Date command go back to 30 days from a particular date

ADate=`date -d"10 days ago" +%s`
BDate=`date -d"$BDate - 30 days" +%s`
It is throwing error as invalid date.

try this:
ADate=`date -d"10 days ago"`
BDate=`date -d"$ADate - 30 days" +%s`
echo $BDate

This works for me:
ADate=$(date -d"10 days ago" +Y%-%m-%d")
BDate=$(date -d"$ADate - 30 days" +%s)
echo $BDate
Output: 1453417200

I have two dates and need to find the difference in hours

Wed Jan 21 20:44:20 EST 2015
Wed Jan 21 19:04:20 EST 2015
I have two dates about, need to get the difference in minutes. Please help
"c=date -d #$(( $(date -d "$b" +%s) - $(date -d "$a" +%s) )) -u +'%H:%M'" -> This command is giving in HH:MM but i want in MM
Thank you

this gives you the result: 100 minutes:
echo $((($(date -d "$a" +%s) - $(date -d "$b" +%s))/60 ))
Note that, it will always give an int value, if you need the precision less than 1 min, like 100.25 you may want to use bc or awk to do the calculation instead of $(( .. ))

How to make date command work in relation to custom specified date?

In Linux there is pretty awesome command date which can be used is ways like this:
# Get some cool date in relation to systems date:
date -d "last Sunday -7 days"
Sun Sep 15 00:00:00 PDT 2013
# Set systems date:
date --set="2013-03-04"
Mon Mar 4 00:00:00 PST 2013
Basically I want to be able to run this command like this:
date --date="last Sunday -7 days" +%Y-%m-%d
2013-09-15
But not in relation to today's system date but in relation to some date generated by another computation in the form of string (e.g. "2013-09-01") or something else.
Please help me to figure out how to do that.

Using a function:
function get_last_day {
local date=$1 day=$2 format=$3 a b i
for (( i = 0; i <= 6; ++i )); do
read -r a b < <(exec date -d "$date - $i days" "+%a $format")
if [[ $a == "$day" ]]; then
echo "$b"
return
fi
done
}
get_last_day '2013-09-18' Sun '%Y-%m-%d'
Output:
2013-09-15

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