In my linux system date is having two space between month and date(ex:Jun 4) .so i used date command(ex:date +'%b %_d %H') and stored the value in variable that time i am getting only one space between month and date(ex:Jun 4), so please help me to get the actual system format while using variable.
date is outputting the string you want; I suspect you aren't properly quoting the variable value when you expand it.
$ d=$(date +'%b %_d %H')
$ echo $d
Jun 6 20
$ echo "$d"
Jun 6 20
Related
I would like to find number of days from two given dates as 25 Aug 2017 and 05 Sep 2017.
My script:
start='2017-08-25';
end='2017-09-05';
ndays=(strtotime($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.
I am trying this simple script but somehow i get 2 days extra each time:
#!/bin/bash
date=$(date +%j)
remaining="$(( 365 - $date))"
echo $remaining
On 12/01/2015 i got 355 days remaining whilst it should be 353.
Date and everything's fine.
The day number is output with a leading zero:
$ date +%j
012
This means that the number will be interpretted as octal, and octal 012 equals decimal 10 (that's why it's off by 2). You can manually specify the radix to override the default interpretation:
date=$(date +%j)
remaining="$(( 365 - 10#$date))"
echo "$remaining"
Gives 353 today, Jan 12.
As a side note, if you want to get the total number of days in the current year instead of hard coding 365, you can use:
days_this_year=$(date -d "$(date +%Y)/12/31" +%j)
I have a shell script that runs on Linux and uses this call to get yesterday's date in YYYY-MM-DD format:
date -d "1 day ago" '+%Y-%m-%d'
It works most of the time, but when the script ran yesterday morning at 2013-03-11 0:35 CDT it returned "2013-03-09" instead of "2013-03-10".
Presumably daylight saving time (which started yesterday) is to blame. I'm guessing the way "1 day ago" is implemented it subtracted 24 hours, and 24 hours before 2013-03-11 0:35 CDT was 2013-03-09 23:35 CST, which led to the result of "2013-03-09".
So what's a good DST-safe way to get yesterday's date in bash on Linux?
I think this should work, irrespective of how often and when you run it ...
date -d "yesterday 13:00" '+%Y-%m-%d'
Under Mac OSX date works slightly different:
For yesterday
date -v-1d +%F
For Last week
date -v-1w +%F
This should also work, but perhaps it is too much:
date -d #$(( $(date +"%s") - 86400)) +"%Y-%m-%d"
If you are certain that the script runs in the first hours of the day, you can simply do
date -d "12 hours ago" '+%Y-%m-%d'
BTW, if the script runs daily at 00:35 (via crontab?) you should ask yourself what will happen if a DST change falls in that hour; the script could not run, or run twice in some cases. Modern implementations of cron are quite clever in this regard, though.
Here a solution that will work with Solaris and AIX as well.
Manipulating the Timezone is possible for changing the clock some hours.
Due to the daylight saving time, 24 hours ago can be today or the day before yesterday.
You are sure that yesterday is 20 or 30 hours ago. Which one? Well, the most recent one that is not today.
echo -e "$(TZ=GMT+30 date +%Y-%m-%d)\n$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
The -e parameter used in the echo command is needed with bash, but will not work with ksh.
In ksh you can use the same command without the -e flag.
When your script will be used in different environments, you can start the script with #!/bin/ksh or #!/bin/bash. You could also replace the \n by a newline:
echo "$(TZ=GMT+30 date +%Y-%m-%d)
$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
date -d "yesterday" '+%Y-%m-%d'
To use this later:
date=$(date -d "yesterday" '+%Y-%m-%d')
you can use
date -d "30 days ago" +"%d/%m/%Y"
to get the date from 30 days ago, similarly you can replace 30 with x amount of days
Just use date and trusty seconds:
As you rightly point out, a lot of the details about the underlying computation are hidden if you rely on English time arithmetic. E.g. -d yesterday, and -d 1 day ago will have different behaviour.
Instead, you can reliably depend on the (precisely documented) seconds since the unix epoch UTC, and bash arithmetic to obtain the moment you want:
date -d #$(( $(date +"%s") - 24*3600)) +"%Y-%m-%d"
This was pointed out in another answer. This form is more portable across platforms with different date command line flags, is language-independent (e.g. "yesterday" vs "hier" in French locale), and frankly (in the long-term) will be easier to remember, because well, you know it already. You might otherwise keep asking yourself: "Was it -d 2 hours ago or -d 2 hour ago again?" or "Is it -d yesterday or -d 1 day ago that I want?"). The only tricky bit here is the #.
Armed with bash and nothing else:
Bash solely on bash, you can also get yesterday's time, via the printf builtin:
%(datefmt)T
causes printf to output the date-time string resulting from using
datefmt as a format string for strftime(3). The corresponding argu‐
ment is an integer representing the number of seconds since the
epoch. Two special argument values may be used: -1 represents the
current time, and -2 represents the time the shell was invoked.
If no argument is specified, conversion behaves as if -1 had
been given.
This is an exception to the usual printf behavior.
So,
# inner printf gets you the current unix time in seconds
# outer printf spits it out according to the format
printf "%(%Y-%m-%d)T\n" $(( $(printf "%(%s)T" -1) - 24*3600 ))
or, equivalently with a temp variable (outer subshell optional, but keeps environment vars clean).
(
now=$(printf "%(%s)T" -1);
printf "%(%Y-%m-%d)T\n" $((now - 24*3600));
)
Note: despite the manpage stating that no argument to the %()T formatter will assume a default -1, i seem to get a 0 instead (thank you, bash manual version 4.3.48)
You can use:
date -d "yesterday 13:55" '+%Y-%m-%d'
Or whatever time you want to retrieve will retrieved by bash.
For month:
date -d "30 days ago" '+%Y-%m-%d'
As this question is tagged bash "DST safe":
And using fork to date command implie delay, there is a simple and more efficient way using pure bash built-in:
printf -v tznow '%(%z %s)T' -1
TZ=${tznow% *} printf -v yesterday '%(%Y-%m-%d)T' $(( ${tznow#* } - 86400 ))
echo $yesterday
This is a lot quicker on more system friendly than having to fork to date.
From bash version 5.0, there is a new variable $EPOCHSECONDS
printf -v tz '%(%z)T' -1
TZ=$tz printf -v yesterday '%(%Y-%m-%d)T' $(( EPOCHSECONDS - 86400 ))
echo $yesterday