我正在尝试从系统中收集信息,我需要以小时和分钟为单位获取当前时间.
目前我有:
date | awk '{print $4}'
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输出如下:
16:18:54
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如何切断秒?
arm*_*ino 89
提供格式字符串:
date +"%H:%M"
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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)
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对于你的问题,我有另一个解决方案。
在第一次使用时date,输出是这样的:
Thu 28 Jan 2021 22:29:40 IST
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然后,如果您只想以小时和分钟显示当前时间,则可以使用以下命令:
date | cut -d " " -f5 | cut -d ":" -f1-2
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然后输出:
22:29
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你可以使用命令
date | awk '{print $4}'| cut -d ':' -f3
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正如您提到的,仅使用date|awk '{print $4}'管道会给您带来类似的结果
20:18:19
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正如我们所看到的,如果我们想要提取该字符串的某些部分,那么我们需要一个分隔符,对于我们的情况来说:,所以我们决定在 的基础上进行切割:。现在这个分隔符会将字符串分成三部分,即 20 ,18 和 19 ,因为我们想要第二部分,所以我们在命令中使用 -f2 。总结 ,
cut:根据分隔符截断一些字符串。
-d:分隔符(此处:)
-f2:我们想要的被砍掉的令牌。
bash 版本 >= 4.2:
printf "%(%H:%M)T\n"
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或者
printf -v foo "%(%H:%M)T\n"
echo "$foo"
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看:man bash
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