Linux 以小时和分钟为单位获取当前时间
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Get current time in hours and minutes
提问by Joe Caruso
I'm trying to collect information from a system and I need to get the current time in hours and minutes.
我正在尝试从系统收集信息，我需要以小时和分钟为单位获取当前时间。
Currently I have:
目前我有：
date | awk '{print }'
which outputs something like:
输出类似：
16:18:54
How do I chop off the seconds?
秒杀怎么办？
采纳答案by armandino
Provide a format string:
提供格式字符串：
date +"%H:%M"
Running man datewill give all the format options
运行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)
回答by Guido
EDIT: whoops. Got it backwards. I though you wanted to only keep the seconds. Edited!
编辑：哎呀。倒过来了。我虽然你只想保留秒数。已编辑！
date +%H:%M
Would be easier, I think :). If you really wanted to chop off the seconds, you could have done
我想会更容易:)。如果你真的想砍掉秒数，你可以做到
date | sed 's/.* \([0-9]*:[0-9]*\):[0-9]*.*//'
回答by Subham Tripathi
you can use
您可以使用
date | awk '{print }'| cut -d ':' -f3
as u mentioned using only date|awk '{print $4}'command gives something like this
正如你提到的，只使用date|awk '{print $4}'命令给出了这样的东西
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 ,
所以我们可以看到如果我们想提取这个字符串的某些部分，那么我们需要一个分隔符，对于我们的例子来说是:这样，所以我们决定基于:. 现在这个分隔符会将字符串分成三部分，即 20 ,18 和 19 ，因为我们想要第二个我们在命令中使用 -f2 。总结 ，
cut: chops some string based on delimeter.
cut: 根据分隔符截取一些字符串。
-d: delimeter (here :)
-d：分隔符（此处:）
-f2: the chopped off token that we want.
-f2：我们想要的切掉的令牌。
