Given a date and time in a format that is not recognized by date, how can I get date to recognize the date and time?
For example:
$ date -d "09SEP2012:23:58:46"
date: invalid date `09SEP2012:23:58:46'
$ date -d "09SEP2012:23:58:46" --magic-option "ddMMMYYY:hh:mm:ss"
Sun Sep 9 23:58:46 MDT 2012
Does --magic-option exist? If not, is there a more elegant way to solve this rather than using sed to transform the input into a well-formed date string?
Neither POSIX nor GNU date have --magic-option. FreeBSD calls it -f (with a syntax similar to date's output format specifiers, not the one you propose).
Your date is very close to being recognized by GNU date: all it takes is replacing the colon that separates the date from the time by a space.
date -d "$(echo "09SEP2012:23:58:46" | sed 's/:/ /')"
01/02/03.
– Gilles 'SO- stop being evil'
Dec 22 '17 at 19:50
I wrote a bunch of tools (dateutils) that deal with dates and times in a more script-friendly way. Your magic option there is --input-format|-i, e.g.:
dconv -i '%d%b%Y:%H:%M:%S' "09SEP2012:23:58:46"
=>
2012-09-09T23:58:46
While dconv does not directly support date's output format (it doesn't confer TZ or anything in the environment), there's a tool strptime in dateutils that does support the %Z format specifier.
If you're using an environment like Ubuntu that has busybox installed, you can do busybox date -D "$input_format" -d "$input", such as:
$ busybox date -D "%m/%d/%Y %H:%M:%S" -d "02/27/2013 23:48:00"
Wed Feb 27 23:48:00 CST 2013
If you want to format the output string to something non-standard you can specify an output format with busybox date -D "$input_format" -d "$input" +"$output_format",
such as
$ busybox date -D "%m/%d/%Y" -d "02/27/2013" +"%d.%-m.%y"
27.2.13
but for ISO-8601-compliant dates, use -I with the level of precision needed, such as:
$ busybox date -D "%m/%d/%Y %H:%M:%S" -d "02/27/2013 23:48:00" -Iseconds
2013-02-27T23:48:00-0600
The busybox man page.
This might not work with %Z timezone format.
