I have read the man pages for date command, but there is no option for converting unix time to string time. I found some websites telling this:
date -d @1343715322
there is no option like @
in the man and info pages. Can someone explain it?
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Sign up to join this communityYou usually won't find details like this in the man pages.
Check the info date
, you will find this in the "Date input formats" section.
It's in the GNU date manual: http://www.gnu.org/software/coreutils/manual/coreutils.html#Seconds-since-the-Epoch. The same explanation is in the coreutils info pages.
If you precede a number with ‘@’, it represents an internal time stamp as a count of seconds. The number can contain an internal decimal point (either ‘.’ or ‘,’); any excess precision not supported by the internal representation is truncated toward minus infinity. Such a number cannot be combined with any other date item, as it specifies a complete time stamp.
Internally, computer times are represented as a count of seconds since an epoch—a well-defined point of time. On GNU and POSIX systems, the epoch is 1970-01-01 00:00:00 utc, so ‘@0’ represents this time, ‘@1’ represents 1970-01-01 00:00:01 utc, and so forth. GNU and most other POSIX-compliant systems support such times as an extension to POSIX, using negative counts, so that ‘@-1’ represents 1969-12-31 23:59:59 utc.
Traditional Unix systems count seconds with 32-bit two's-complement integers and can represent times from 1901-12-13 20:45:52 through 2038-01-19 03:14:07 utc. More modern systems use 64-bit counts of seconds with nanosecond subcounts, and can represent all the times in the known lifetime of the universe to a resolution of 1 nanosecond.
On most hosts, these counts ignore the presence of leap seconds. For example, on most hosts ‘@915148799’ represents 1998-12-31 23:59:59 utc, ‘@915148800’ represents 1999-01-01 00:00:00 utc, and there is no way to represent the intervening leap second 1998-12-31 23:59:60 utc.
date
implementation you have?@
is specific for GNUdate
.@
but still uses it in an example. Yet I don't understand your question.@
seems to be the only way to do this withdate
so where's the problem?