View Full Version : [SOLVED] Eliminate leading zero in Bash

September 2nd, 2012, 09:47 PM

I am using bash and need to eliminate the leading zero in the current month: $( date +'%m')
For example, now is September, and I need it to give me '9', not '09', so I can use the month in calculations.

Can anyone please help?


September 2nd, 2012, 09:57 PM
~$ date +'%-m'
~$ date --help
Usage: date [OPTION]... [+FORMAT]
or: date [-u|--utc|--universal] [MMDDhhmm[[CC]YY][.ss]]
Display the current time in the given FORMAT, or set the system date.

-d, --date=STRING display time described by STRING, not `now'
-f, --file=DATEFILE like --date once for each line of DATEFILE
-r, --reference=FILE display the last modification time of FILE
-R, --rfc-2822 output date and time in RFC 2822 format.
Example: Mon, 07 Aug 2006 12:34:56 -0600
--rfc-3339=TIMESPEC output date and time in RFC 3339 format.
TIMESPEC=`date', `seconds', or `ns' for
date and time to the indicated precision.
Date and time components are separated by
a single space: 2006-08-07 12:34:56-06:00
-s, --set=STRING set time described by STRING
-u, --utc, --universal print or set Coordinated Universal Time
--help display this help and exit
--version output version information and exit

FORMAT controls the output. Interpreted sequences are:

%% a literal %
%a locale's abbreviated weekday name (e.g., Sun)
%A locale's full weekday name (e.g., Sunday)
%b locale's abbreviated month name (e.g., Jan)
%B locale's full month name (e.g., January)
%c locale's date and time (e.g., Thu Mar 3 23:05:25 2005)
%C century; like %Y, except omit last two digits (e.g., 20)
%d day of month (e.g., 01)
%D date; same as %m/%d/%y
%e day of month, space padded; same as %_d
%F full date; same as %Y-%m-%d
%g last two digits of year of ISO week number (see %G)
%G year of ISO week number (see %V); normally useful only with %V
%h same as %b
%H hour (00..23)
%I hour (01..12)
%j day of year (001..366)
%k hour, space padded ( 0..23); same as %_H
%l hour, space padded ( 1..12); same as %_I
%m month (01..12)
%M minute (00..59)
%n a newline
%N nanoseconds (000000000..999999999)
%p locale's equivalent of either AM or PM; blank if not known
%P like %p, but lower case
%r locale's 12-hour clock time (e.g., 11:11:04 PM)
%R 24-hour hour and minute; same as %H:%M
%s seconds since 1970-01-01 00:00:00 UTC
%S second (00..60)
%t a tab
%T time; same as %H:%M:%S
%u day of week (1..7); 1 is Monday
%U week number of year, with Sunday as first day of week (00..53)
%V ISO week number, with Monday as first day of week (01..53)
%w day of week (0..6); 0 is Sunday
%W week number of year, with Monday as first day of week (00..53)
%x locale's date representation (e.g., 12/31/99)
%X locale's time representation (e.g., 23:13:48)
%y last two digits of year (00..99)
%Y year
%z +hhmm numeric time zone (e.g., -0400)
%:z +hh:mm numeric time zone (e.g., -04:00)
%::z +hh:mm:ss numeric time zone (e.g., -04:00:00)
%:::z numeric time zone with : to necessary precision (e.g., -04, +05:30)
%Z alphabetic time zone abbreviation (e.g., EDT)

By default, date pads numeric fields with zeroes.
The following optional flags may follow `%':

- (hyphen) do not pad the field
_ (underscore) pad with spaces
0 (zero) pad with zeros
^ use upper case if possible
# use opposite case if possible

After any flags comes an optional field width, as a decimal number;
then an optional modifier, which is either
E to use the locale's alternate representations if available, or
O to use the locale's alternate numeric symbols if available.

Convert seconds since the epoch (1970-01-01 UTC) to a date
$ date --date='@2147483647'

Show the time on the west coast of the US (use tzselect(1) to find TZ)
$ TZ='America/Los_Angeles' date

Show the local time for 9AM next Friday on the west coast of the US
$ date --date='TZ="America/Los_Angeles" 09:00 next Fri'

Report date bugs to bug-coreutils@gnu.org
GNU coreutils home page: <http://www.gnu.org/software/coreutils/>
General help using GNU software: <http://www.gnu.org/gethelp/>
For complete documentation, run: info coreutils 'date invocation'

September 2nd, 2012, 10:06 PM
post above solves the problem but here is some trivia which may be useful in other scenarios
leading zeros make bash assume the number is in base 8 system.
In that case you can force it to use base 10

$ x=09; echo $(( 10#$x+12 ))

btw, what kind of calculations do you need? because it's possible you are reinventing the wheel.

September 2nd, 2012, 10:17 PM
It works! Thank you guys for your help.