How to convert 2019-08-15 date format to August 15, 2019 in the command line?Convert Date formate in...
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How to convert 2019-08-15 date format to August 15, 2019 in the command line?
Convert Date formate in unixConvert any Date format in unixconvert date format in logConvert a date formatchange date format and store in variable using awkHow to convert date format in fileDate change format in unixCan I convert a date in the format YYYYMMDDHHMM using date?convert the dates to a standard format
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Given a bash
variable with the value 2019-08-15
, is there some utility that can convert that date to the format August 15, 2019
?
command-line date
add a comment |
Given a bash
variable with the value 2019-08-15
, is there some utility that can convert that date to the format August 15, 2019
?
command-line date
1
Possible duplicate of Convert Date formate in unix
– muru
yesterday
add a comment |
Given a bash
variable with the value 2019-08-15
, is there some utility that can convert that date to the format August 15, 2019
?
command-line date
Given a bash
variable with the value 2019-08-15
, is there some utility that can convert that date to the format August 15, 2019
?
command-line date
command-line date
edited yesterday
Kusalananda♦
160k18 gold badges316 silver badges502 bronze badges
160k18 gold badges316 silver badges502 bronze badges
asked yesterday
VillageVillage
2,1558 gold badges35 silver badges57 bronze badges
2,1558 gold badges35 silver badges57 bronze badges
1
Possible duplicate of Convert Date formate in unix
– muru
yesterday
add a comment |
1
Possible duplicate of Convert Date formate in unix
– muru
yesterday
1
1
Possible duplicate of Convert Date formate in unix
– muru
yesterday
Possible duplicate of Convert Date formate in unix
– muru
yesterday
add a comment |
2 Answers
2
active
oldest
votes
On Linux, or any system that uses GNU date
:
$ thedate=2019-08-15
$ date -d "$thedate" +'%B %e, %Y'
August 15, 2019
On macOS, OpenBSD and FreeBSD, where GNU date
is not available by default:
$ thedate=2019-08-15
$ date -j -f '%Y-%m-%d' "$thedate" +'%B %e, %Y'
August 15, 2019
The -j
option disables setting the system clock, and the format string used with -f
describes the input date format (should be a strptime(3)
format string describing the format used by your variable's value). Then follows the value of your variable and the format that you want your output to be in (should be a strftime(3)
format string).
NetBSD users may use something similar to the above but without the -f input_fmt
option, as their date
implementation uses parsedate(3)
. Note also the -d
option to specify the input date string:
$ thedate=2019-08-15
$ date -j -d "$thedate" +'%B %e, %Y'
August 15, 2019
See also the manual for date
on your system.
add a comment |
Assuming that you have access to GNU date
, something along
$ date --date="2019-08-15" "+%B %d, %Y"
August 15, 2019
Check the manpage of date (man date
).
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
On Linux, or any system that uses GNU date
:
$ thedate=2019-08-15
$ date -d "$thedate" +'%B %e, %Y'
August 15, 2019
On macOS, OpenBSD and FreeBSD, where GNU date
is not available by default:
$ thedate=2019-08-15
$ date -j -f '%Y-%m-%d' "$thedate" +'%B %e, %Y'
August 15, 2019
The -j
option disables setting the system clock, and the format string used with -f
describes the input date format (should be a strptime(3)
format string describing the format used by your variable's value). Then follows the value of your variable and the format that you want your output to be in (should be a strftime(3)
format string).
NetBSD users may use something similar to the above but without the -f input_fmt
option, as their date
implementation uses parsedate(3)
. Note also the -d
option to specify the input date string:
$ thedate=2019-08-15
$ date -j -d "$thedate" +'%B %e, %Y'
August 15, 2019
See also the manual for date
on your system.
add a comment |
On Linux, or any system that uses GNU date
:
$ thedate=2019-08-15
$ date -d "$thedate" +'%B %e, %Y'
August 15, 2019
On macOS, OpenBSD and FreeBSD, where GNU date
is not available by default:
$ thedate=2019-08-15
$ date -j -f '%Y-%m-%d' "$thedate" +'%B %e, %Y'
August 15, 2019
The -j
option disables setting the system clock, and the format string used with -f
describes the input date format (should be a strptime(3)
format string describing the format used by your variable's value). Then follows the value of your variable and the format that you want your output to be in (should be a strftime(3)
format string).
NetBSD users may use something similar to the above but without the -f input_fmt
option, as their date
implementation uses parsedate(3)
. Note also the -d
option to specify the input date string:
$ thedate=2019-08-15
$ date -j -d "$thedate" +'%B %e, %Y'
August 15, 2019
See also the manual for date
on your system.
add a comment |
On Linux, or any system that uses GNU date
:
$ thedate=2019-08-15
$ date -d "$thedate" +'%B %e, %Y'
August 15, 2019
On macOS, OpenBSD and FreeBSD, where GNU date
is not available by default:
$ thedate=2019-08-15
$ date -j -f '%Y-%m-%d' "$thedate" +'%B %e, %Y'
August 15, 2019
The -j
option disables setting the system clock, and the format string used with -f
describes the input date format (should be a strptime(3)
format string describing the format used by your variable's value). Then follows the value of your variable and the format that you want your output to be in (should be a strftime(3)
format string).
NetBSD users may use something similar to the above but without the -f input_fmt
option, as their date
implementation uses parsedate(3)
. Note also the -d
option to specify the input date string:
$ thedate=2019-08-15
$ date -j -d "$thedate" +'%B %e, %Y'
August 15, 2019
See also the manual for date
on your system.
On Linux, or any system that uses GNU date
:
$ thedate=2019-08-15
$ date -d "$thedate" +'%B %e, %Y'
August 15, 2019
On macOS, OpenBSD and FreeBSD, where GNU date
is not available by default:
$ thedate=2019-08-15
$ date -j -f '%Y-%m-%d' "$thedate" +'%B %e, %Y'
August 15, 2019
The -j
option disables setting the system clock, and the format string used with -f
describes the input date format (should be a strptime(3)
format string describing the format used by your variable's value). Then follows the value of your variable and the format that you want your output to be in (should be a strftime(3)
format string).
NetBSD users may use something similar to the above but without the -f input_fmt
option, as their date
implementation uses parsedate(3)
. Note also the -d
option to specify the input date string:
$ thedate=2019-08-15
$ date -j -d "$thedate" +'%B %e, %Y'
August 15, 2019
See also the manual for date
on your system.
edited yesterday
answered yesterday
Kusalananda♦Kusalananda
160k18 gold badges316 silver badges502 bronze badges
160k18 gold badges316 silver badges502 bronze badges
add a comment |
add a comment |
Assuming that you have access to GNU date
, something along
$ date --date="2019-08-15" "+%B %d, %Y"
August 15, 2019
Check the manpage of date (man date
).
add a comment |
Assuming that you have access to GNU date
, something along
$ date --date="2019-08-15" "+%B %d, %Y"
August 15, 2019
Check the manpage of date (man date
).
add a comment |
Assuming that you have access to GNU date
, something along
$ date --date="2019-08-15" "+%B %d, %Y"
August 15, 2019
Check the manpage of date (man date
).
Assuming that you have access to GNU date
, something along
$ date --date="2019-08-15" "+%B %d, %Y"
August 15, 2019
Check the manpage of date (man date
).
edited yesterday
Kusalananda♦
160k18 gold badges316 silver badges502 bronze badges
160k18 gold badges316 silver badges502 bronze badges
answered yesterday
JankaJanka
2965 bronze badges
2965 bronze badges
add a comment |
add a comment |
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1
Possible duplicate of Convert Date formate in unix
– muru
yesterday