以小时和分钟表示当前时间

我正试图从一个系统中收集信息,我需要得到当前时间的小时和分钟。

目前我有:

date | awk '{print $4}'

它的输出类似于:

16:18:54

我要怎么把秒数剪掉?

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Provide a format string:

date +"%H:%M"

Running man date will give all the format options

%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)
date +%H:%M

Would be easier, I think :). If you really wanted to chop off the seconds, you could have done

date | sed 's/.* \([0-9]*:[0-9]*\):[0-9]*.*/\1/'

you can use command

date | awk '{print $4}'| cut -d ':' -f3

as you mentioned using only the date|awk '{print $4}' pipeline gives you something like this

20:18:19

so as we can see if we want to extract some part of this string then we need a delimiter , for our case it is :, so we decide to chop on the basis of :. Now this delimiter will chop the string into three parts i.e. 20 ,18 and 19 , as we want the second one we use -f2 in our command. to sum up ,

cut : chops some string based on delimeter.

-d : delimeter (here :)

-f2 : the chopped off token that we want.

Could also potentially use this script to use the system time in a variable

now=$(date +"%m_%d_%Y_%M:%S")

Which outputs as

12_07_2020_34:21

With bash version >= 4.2:

printf "%(%H:%M)T\n"

or

printf -v foo "%(%H:%M)T\n"
echo "$foo"

See: man bash

I have another solution for your question .

In the first when use date the output is like this :

Thu 28 Jan 2021 22:29:40 IST

Then if you want only to show current time in hours and minutes you can use this command :

date | cut -d " " -f5 | cut -d ":" -f1-2

Then the output :

22:29