I'm trying to convert a string into seconds using date +%s, but everytime i get this error:
deadline='Monday August 2 10:45:25 AM GMT'
date -d $deadline +%s
date: extra operand'2'
But if I don't use a variable this command works fine
date -d 'Monday August 2 10:45:25 AM GMT' +%s
1627901012
Can somebody explain why this command doesn't work with my variable?
why this command doesn't work with my variable?
Unquoted variable expansions undergo word splitting.
After expansion, you are executing:
date -d Monday August 2 10:45:25 AM GMT +%s
Monday is the argument for -d, August is the one argument that date takes. Then the rest August 2 10:... are just invalid positional arguments for date, so the tool exits with an error that too many arguments were passed.
Related
I am trying to add hours to a given datetime in bash and its giving weird results
This is just to check if its parsing the date time correctly
$ date -d "2020-01-21 12:00:00"
Tue 21 Jan 12:00:00 UTC 2020
Now when I try to do date math
$ date -d "2020-01-21 12:00:00 +2 hour"
Tue 21 Jan 11:00:00 UTC 2020
I tried few other operations but similar behaviour.
If I change my format then it behaves correctly for eg.
date -d "12:00:00 2020-01-21 +2 hour"
Tue 21 Jan 14:00:00 UTC 2020
not sure whats going on here.
The +2 is being interpreted as an explicit timezone specifier (equivalent to EET), so date -d "2020-01-21 12:00:00 +2 hour" is interpreted the same as date -d "2020-01-21 12:00:00 EET hour", which adds one hour to the timezone-adjusted time specified.
You can either provide an explicit timezone (as suggested by Maaz) so that +2 hour is syntactically an additional offset, or you can move it to the beginning of the expression, where it can't be parsed as a timezone.
date -d "+ 2 hour 2020-01-21 12:00:00"
Basically you need to pass the timezone in command as well like UTC/CT/IST, so that date command can understand the point of reference to add extra hours.
For example if you are in UTC timezone, then following will produce correct output for you.
date -d "2020-01-21 12:00:00 UTC + 2 hour"
How can I interpolate a bash command in a Heroku scheduler command?
I have a command that runs everyday & it takes a day. Now I want to make it dynamic using bash date command e.g
cli "$(date --date "7 day ago")"
For today it will be Sat Apr 27 22:36:46 +06 2019 and for tomorrow it will be Sun Apr 28 22:36:46 +06 2019.
How can I achieve this?
Stack Overflow's syntax highlighting makes the problem pretty clear. You're nesting double quotes inside double quotes without escaping them:
cli "$(date --date "7 day ago")"
This is interpreted as three arguments:
"$(date --date "7
day
ago")"
Replace the innner ones with single quotes and it should work:
cli "$(date --date '7 day ago')"
Another alternative would be to escape the inner quotes with backslashes, but IMO using single quotes is more readable.
I would like to find number of days from two given dates as 25 Aug 2017 and 05 Sep 2017.
My script:
start='2017-08-25';
end='2017-09-05';
ndays=(strtotime($end)- strtotime($start))/24/3600;
echo $ndays
When run this script, I am getting following error messages.
Line 3: syntax error near unexpected token `('
Desire output value:
10
No shell that I am aware of has any tools for working with dates. At the very least, you need an external tool like date to convert your dates to an intermediate form, like the number of seconds since some fixed point in time (Unix uses Jan 1, 1970), then do your calculation with those values before processing the result further.
Assuming the use of GNU date from the Linux tag, you would do something like
start='2017-08-25'
end='2017-09-05'
start_seconds=$(date +%s --date "$start" --utc)
end_seconds=$(date +%s --date "$end" --utc)
ndays=$(( (end_seconds - start_seconds) / 24 / 3600 ));
echo $ndays
Note that since most shells only support integer arithmetic, this won't give you an exact number of days.
You can use date to convert each date to epoch seconds, then use arithmetic expansion to do the subtraction and conversion back to days:
#! /bin/bash
start='2017-08-25';
end='2017-09-05';
diff=$(date -d $end +%s)-$(date -d $start +%s)
echo $(( ($diff) / 60 / 60 / 24 )) # 11
I would like to get the current and next months using shell script, I have tried this command:
$ date '+%b'
mar
$ date +"%B %Y" --date="$(date +%Y-%m-15) next month"
March 2018
But it always displays only the current month.
Could you please help me if there is something wrong with the commands.
$ date -d "next month" '+%B %Y'
April 2018
Check this post about specific caveats
Note: your command works just fine for me (archlinux, bash4, date GNU coreutils 8.29)
I wouldn't rely on date alone to do this. Instead, perform a little basic math on the month number.
this_month=$(date +%-m) # GNU extension to avoid leading 0
next_month=$(( this_month % 12 + 1 ))
next_month_name=$(date +%B --date "2018-$next_month-1")
Since you are using bash, you don't need to use date at all to get the current month; the built-in printf can call the underlying date/time routines itself, saving a fork.
$ printf -v this_month '%(%-m)T\n'
$ echo $this_month
3
What variant and version of date are you running? "Solaris 5.2" was never released, though SunOS 5.2 was a kernel in Solaris 2.2 (EOL in 1999). See the Solaris OS version history. The Solaris 10 (SunOS 5.10) man page for date does not support GNU's --date= syntax used in the question, so I'm guessing you're using some version of date from GNU coreutils.
Here's a solution using BSD date (note, this is academic in the face of BSD's date -v 1m):
date -jf %s $((1728000+$(date -jf %Y-%m-%d $(date +%Y-%m-15) +%s))) +"%B %Y"
There are three date calls here. BSD's date allows specifying a format (GNU can intuit most formats on its own). The parent call is the one that takes the final time (as seconds since the 1970 epoch), expressing it in the desired "Month Year" format. Seconds are determine by adding 20 days to the epoch time of the current month on the 15th day. Since no month has 15+20 days, this is always the following month.
Here's a direct translation of that logic to GNU date:
date +"%B %Y" --date="#$((1728000+$(date +%s --date=$(date +%Y-%m-15))))"
Here's a simpler solution using GNU date, with one fewer date call:
date +"%B %Y" --date="$(date +%Y-%m-15) 20 days"
(A bug in GNU date will give you the wrong month if you run date --date="next month" on the 31st.)
The below one works in Red Hat 4.4.7-23, Linux version 2.6.32-754.2.1.el6.x86_64.
Just use the "month" for future months and "month ago" for previous months.. Dont confuse with adding +/- signs to the number. Check out.
> date "+%B-%Y" #current month
November-2018
> date -d" 1 month" "+%B-%Y"
December-2018
> date -d" 1 month ago" "+%B-%Y"
October-2018
>
More..
> date -d" 7 month ago" "+%B-%Y"
April-2018
> date -d" 7 month " "+%B-%Y"
June-2019
>
In my Unix shell programming, I am trying to get the next date (tomorrow's date) over a reference date defined as "a". Here is the code:
a=2016-01-02
Which operator would I use in my code so that Unix will automatically define a as tomorrow's date as in below
a=2016-01-03
date has a -d option that is very useful in this situation.
To get the next day, add a space after the date then add 1 day
date +%Y-%m-%d -d "$a 1 day"
It's important to add the format specifier because without it, you would get the following output
=>"Sun Jan 3 00:00:00 UTC 2016"
To update the a variable, you could do something like this
a=$(date +%Y-%m-%d -d "$a 1 day")
Remember to wrap the command inside of parentheses with a $ sign in front of it.