Hi I have a bash script called by a systemd service that needs to run until the start of the next hour. Currently I have been using:
currentTime=$(date +"%s")
nextHour=$(date -d "$(date -d 'next hour' '+%H:00:00')" '+%s')
duration=$(((nextHour-currentTime)*1000))
Which works except for trying to calculate the difference between 11pm and midnight were as far as I can tell it gets the current days midnight from 23 hours previous.
Oct 13 23:00:05 host bash[2019]: 1665698405
Oct 13 23:00:05 host bash[2019]: 1665615600
Oct 13 23:00:05 host bash[2019]: -82805000
I figure I could put a conditional check with a different calculation if needed or perhaps look at a systemd timer for triggering the service but as the service needs to always activate on boot/reboot as well as running hour to hour this setup seemed more appropriate.
Would appreciate any advice on why this is happening and advice on most streamlined steps to avoid it.
This isn't really a question about bash, but seems to be more about date. Given that date has multiple different implementations, it seems wiser to choose a different tool to do the calculation. I suspect perl is more standardized and (almost) as available as date, so you might try to get the difference with:
perl -MTime::Seconds -MTime::Piece -E '
my $t = localtime; my $m = ($t + ONE_HOUR)->truncate(to => "hour"); say $m - $t'
I want to validate, the date given as input is 7 day old or more from current day in bash script linux
This command provides the date 7 days previous to now, in seconds (since 1/1/1970):
minus7=$(date -d "7 days ago" "+%s")
This command provides the date in seconds of the user's input date:
tstDate=$(date -d "$yourInputDate" "+%s")
then
let delta=$tstDate-$minus7
i need output with Year in start field, below command i am using in Linux to get the License details but i am getting date like(start Tue 1/7 9:54) , so want to know the year for this:
lmutil lmstat -a -c user#server_name -f abcd -t
Output : start Tue 1/7 9:54
Looking for start Tue 1/7(with year) 9:54
Please help.
Thanks
I want to run command with specific date in history.
Let say today is 12.Nov.2019 and I need to run one command with date of 10th. Is there any possibility of setting date on fly ?
date -d '1 day ago' /path/to/command
doesn't work.
Use can you faketime to run a command with given time:
faketime - manipulate the system time for a given command
Because you didn't tell us what command and time you want to run,
I suppose that you want to run /path/to/command on Oct 10th, 2019.
Run /path/to/command in 2 days ago:
faketime -f -2d /path/to/command
Run /path/to/command on Oct 10th, 2019:
faketime -f '2019-11-10' /path/to/command
I am running SunOS.
bash-3.00$ uname -a
SunOS lvsaishdc3in0001 5.10 Generic_142901-02 i86pc i386 i86pc
I need to find Yesterday's date in linux with the proper formatting passed from command prompt. When I tried like this on my shell prompt-
bash-3.00$ date --date='yesterday' '+%Y%m%d'
date: illegal option -- date=yesterday
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
I always get date illegal option, why is it so?
Is there anything wrong I am doing?
Update:-
bash-3.00$ date --version
date: illegal option -- version
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
Try this below thing. It should work
YESTERDAY=`TZ=GMT+24 date +%Y%m%d`; echo $YESTERDAY
Try this one out:
DATE_STAMP=`TZ=GMT+24 date +%Y%m%d`
where GMT is the time zone and you might need to alter the 24 according to the hours difference you have from GMT. Either that or you can change GMT to a time zone more comfortable to you e.g. CST
As larsks suggested, you can use perl:
perl -e 'use POSIX qw(strftime); print strftime "%a %b %e %H:%M:%S %Y",localtime(time()- 3600*24);'
Slightly modified from
http://blog.rootshell.be/2006/05/04/solaris-yesterday-date/
To get YYYYMMDD format use this
perl -e 'use POSIX qw(strftime); print strftime "%Y%m%d",localtime(time()- 3600*24);'
This link explains how to format date and time with strftime
http://perltraining.com.au/tips/2009-02-26.html
A pure bash solution
#!/bin/bash
# get and split date
today=`date +%Y%m%d`
year=${today:0:4}
month=${today:4:2}
day=${today:6:2}
# avoid octal mismatch
if (( ${day:0:1} == 0 )); then day=${day:1:1}; fi
if (( ${month:0:1} == 0 )); then month=${month:1:1}; fi
# calc
day=$((day-1))
if ((day==0)); then
month=$((month-1))
if ((month==0)); then
year=$((year-1))
month=12
fi
last_day_of_month=$((((62648012>>month*2&3)+28)+(month==2 && y%4==0)))
day=$last_day_of_month
fi
# format result
if ((day<10)); then day="0"$day; fi
if ((month<10)); then month="0"$month; fi
yesterday="$year$month$day"
echo $yesterday
TZ=$TZ+24 date +'%Y/%m/%d' in SunOS.
Playing on Solaris10 with non-GMT environment, I'm getting this:
# date
Fri Jul 26 13:09:38 CEST 2013 (OK)
# (TZ=CEST+24 date)
Thu Jul 25 11:09:38 CEST 2013 (ERR)
# (TZ=GMT+24 date)
Thu Jul 25 11:09:38 GMT 2013 (OK)
# (TZ=CEST+$((24-$((`date "+%H"`-`date -u "+%H"`)))) date)
Thu Jul 25 13:09:38 CEST 2013 (OK)
As You colud see, I have and I want to get CEST , but TZ=CEST+24 giving me wrong CEST data; GMT+24 giving me correct data, but unusable.
To get the proper result, I has to use GMT+22 (wrong command, correct result) or CEST+22 (wrong value, but finnaly correct result for correct TZ)
A pure bash solution given by #olivecoder is very reliable compared to any other solution but there is a mistake to be corrected. when the day fall on 1st of the month the script is failing with date saying "last_day_of_month" in day value. #olivecoder has missed $ in
day=last_day_of_month, that it should be
day=$last_day_of_month;
After this correction it works very good.
Using Timezone -24 is having some issue based on time when use it. in some cases it goes to day before yesterday. So I think its not reliable.