以小时和分钟获取当前时间

I'm trying to collect information from a system and I need to get the current time in hours and minutes.

我正在尝试从系统中收集信息，我需要以小时和分钟的形式获取当前时间。

Currently I have:

目前我有：

date | awk '{print $4}'

which outputs something like:

输出如下：

16:18:54

How do I chop off the seconds?

如何切断秒？

3 个解决方案

#1

Provide a format string:

提供格式字符串：

date +"%H:%M"

Running man date will give all the format options

Running man date将提供所有格式选项

%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)

#2

you can use

您可以使用

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

as u mentioned using only date|awk '{print $4}' command gives something like this

正如你所提到的那样只使用日期| awk'{print $ 4}'命令给出了类似的东西

20:18:19

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

因此，我们可以看到，如果我们想要提取此字符串的某些部分，那么我们需要一个分隔符，对于我们的情况它是：所以我们决定在以下基础上进行切割：。现在这个分隔符会将字符串分成三个部分，即20,18和19，因为我们想要在命令中使用第二个-f2。总结一下，

cut : chops some string based on delimeter.

cut：根据deimeter切换一些字符串。

-d : delimeter (here :)

-d：delimeter（这里:)

-f2 : the chopped off token that we want.

-f2：我们想要的切断标记。

#3

EDIT: whoops. Got it backwards. I though you wanted to only keep the seconds. Edited!

编辑：哎呀。落后了。我想你只想保持秒数。编辑！

date +%H:%M

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

会更容易，我想:)。如果你真的想要砍掉秒，你就可以做到

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

#1