Convert a time span in seconds to formatted time i

问题:

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?

回答1:

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

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

Explanation:

• The date utility 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 -u option 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 -d part tells date to accept the time information from string instead of using now
  • The @${i} part is how you tell date that $i is in seconds
• The +"%T" is for formatting your output. From the man date page: %T time; same as %H:%M:%S. Since we only care about the HH:MM:SS part, this fits!

回答2:

Another approach: arithmetic

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

回答3:

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

In BSD/OS X, use

date -u -r $i +%T

回答4:

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

convertsecs() {
 ((h=${1}/3600))
 ((m=(${1}%3600)/60))
 ((s=${1}%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:

# convert seconds to day-hour:min:sec
convertsecs2dhms() {
 ((d=${1}/(60*60*24)))
 ((h=(${1}%(60*60*24))/(60*60)))
 ((m=(${1}%(60*60))/60))
 ((s=${1}%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5:

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

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

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

回答6:

I use C shell, like this:

#! /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7:

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 @$i +%T or in some cases date -d @"$i" +%T.