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/:/ /')"
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    and with custom formats? Only way to go are regular expressions? this is horrible :-( – phil294 Dec 22 '17 at 19:39
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    You can't parse dates without knowing something about the format they're in, the classical example being 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"

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.

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