Linux 将时间跨度（以秒为单位）转换为 shell 中的格式化时间

Convert a time span in seconds to formatted time in shell

提问by Darren

I have a variable of $i which is seconds in a shell script, and I am trying to convert it to 24 HOUR HH:MM:SS. Is this possible in shell?

我有一个 $i 变量，它是 shell 脚本中的秒数，我试图将它转换为 24 小时 HH:MM:SS。这在shell中可能吗？

采纳答案by sampson-chen

Here's a fun hacky way to do exactly what you are looking for =)

这是一个有趣的 hacky 方式来做你正在寻找的东西 =)

date -u -d @${i} +"%T"

Explanation:

解释：

• The dateutility allows you to specify a time, from string, in seconds since 1970-01-01 00:00:00 UTC, and output it in whatever format you specify.
• The -uoption is to display UTC time, so it doesn't factor in timezone offsets (since start time from 1970 is in UTC)
• The following parts are GNU date-specific (Linux):
  • The -dpart tells dateto accept the time information from string instead of using now
  • The @${i}part is how you tell datethat $iis in seconds
• The +"%T"is for formatting your output. From the man datepage: %T time; same as %H:%M:%S. Since we only care about the HH:MM:SSpart, this fits!

• 该date实用程序允许您从字符串中指定自 1970-01-01 00:00:00 UTC 以来的秒数，并以您指定的任何格式输出。
• 该-u选项是显示 UTC 时间，因此它不考虑时区偏移（因为 1970 年的开始时间是 UTC）
• 以下部分是 GNUdate特定的 (Linux)：
  • 该-d部分告诉date接受来自字符串的时间信息而不是使用now
  • 该@${i}部分是你如何告诉date那$i是在几秒钟内
• 该+"%T"是格式化你的输出。从man date页面：%T time; same as %H:%M:%S。由于我们只关心HH:MM:SS零件，所以这很合适！

回答by Basile Starynkevitch

If $irepresents some date in second since the Epoch, you could display it with

如果$i代表自纪元以来的某个日期，您可以使用

  date -u -d @$i +%H:%M:%S

but you seems to suppose that $iis an interval (e.g. some duration) not a date, and then I don't understand what you want.

但您似乎认为这$i是一个间隔（例如某个持续时间）而不是日期，然后我不明白您想要什么。

回答by glenn Hymanman

Another approach: arithmetic

另一种方法：算术

i=6789
((sec=i%60, i/=60, min=i%60, hrs=i/60))
timestamp=$(printf "%d:%02d:%02d" $hrs $min $sec)
echo $timestamp

produces 1:53:09

产生 1:53:09

回答by Alan Tam

The -dargument applies to date from coreutils(Linux) only.

该-d参数仅适用于coreutils(Linux) 的日期。

In BSD/OS X, use

在 BSD/OS X 中，使用

date -u -r $i +%T

回答by kossboss

Here is my algo/script helpers on my site: http://ram.kossboss.com/seconds-to-split-time-convert/I used this elogant algo from here: Convert seconds to hours, minutes, seconds

这是我网站上的算法/脚本助手： http://ram.kossboss.com/seconds-to-split-time-convert/ 我从这里使用了这个优雅的算法：将秒转换为小时、分钟、秒

convertsecs() {
 ((h=/3600))
 ((m=(%3600)/60))
 ((s=%60))
 printf "%02d:%02d:%02d\n" $h $m $s
}
TIME1="36"
TIME2="1036"
TIME3="91925"

echo $(convertsecs $TIME1)
echo $(convertsecs $TIME2)
echo $(convertsecs $TIME3)

Example of my second to day, hour, minute, second converter:

我的秒到天、小时、分钟、秒转换器的示例：

# convert seconds to day-hour:min:sec
convertsecs2dhms() {
 ((d=/(60*60*24)))
 ((h=(%(60*60*24))/(60*60)))
 ((m=(%(60*60))/60))
 ((s=%60))
 printf "%02d-%02d:%02d:%02d\n" $d $h $m $s
 # PRETTY OUTPUT: uncomment below printf and comment out above printf if you want prettier output
 # printf "%02dd %02dh %02dm %02ds\n" $d $h $m $s
}
# setting test variables: testing some constant variables & evaluated variables
TIME1="36"
TIME2="1036"
TIME3="91925"
# one way to output results
((TIME4=$TIME3*2)) # 183850
((TIME5=$TIME3*$TIME1)) # 3309300
((TIME6=100*86400+3*3600+40*60+31)) # 8653231 s = 100 days + 3 hours + 40 min + 31 sec
# outputting results: another way to show results (via echo & command substitution with         backticks)
echo $TIME1 - `convertsecs2dhms $TIME1`
echo $TIME2 - `convertsecs2dhms $TIME2`
echo $TIME3 - `convertsecs2dhms $TIME3`
echo $TIME4 - `convertsecs2dhms $TIME4`
echo $TIME5 - `convertsecs2dhms $TIME5`
echo $TIME6 - `convertsecs2dhms $TIME6`

# OUTPUT WOULD BE LIKE THIS (If none pretty printf used): 
# 36 - 00-00:00:36
# 1036 - 00-00:17:16
# 91925 - 01-01:32:05
# 183850 - 02-03:04:10
# 3309300 - 38-07:15:00
# 8653231 - 100-03:40:31
# OUTPUT WOULD BE LIKE THIS (If pretty printf used): 
# 36 - 00d 00h 00m 36s
# 1036 - 00d 00h 17m 16s
# 91925 - 01d 01h 32m 05s
# 183850 - 02d 03h 04m 10s
# 3309300 - 38d 07h 15m 00s
# 1000000000 - 11574d 01h 46m 40s

回答by Shilv

I use C shell, like this:

我使用 C shell，如下所示：

#! /bin/csh -f

set begDate_r = `date +%s`
set endDate_r = `date +%s`

set secs = `echo "$endDate_r - $begDate_r" | bc`
set h = `echo $secs/3600 | bc`
set m = `echo "$secs/60 - 60*$h" | bc`
set s = `echo $secs%60 | bc`

echo "Formatted Time: $h HOUR(s) - $m MIN(s) - $s SEC(s)"

回答by Shai Alon

Continuing @Daren`s answer, just to be clear: If you want to use the conversion to your time zone, don't use the "u" switch, as in: date -d @$i +%Tor in some cases date -d @"$i" +%T.

继续@Daren 的回答，只是要清楚：如果您想使用时区的转换，请不要使用“u”开关，例如：date -d @$i +%T或在某些情况下date -d @"$i" +%T。