How to calculate time elapsed in bash script

Question

I print the start and end time using date    T   which results in something like   10 33 56 10 36 10   How could I calculate and print the difference between these two   I would like to get something like   2m 14s

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Another option is to use datediff from dateutils  http   www fresse org dateutils  datediff      datediff 10 33 56 10 36 10 134s   datediff 10 33 56 10 36 10 -f H  M  S 0 2 14   datediff 10 33 56 10 36 10 -f 0H  0M  0S 00 02 14   You could also use gawk  mawk 1 3 4 also has strftime and mktime but older versions of mawk and nawk don t     TZ UTC0 awk  BEGIN print strftime   T  mktime  1970 1 1 10 36 10  -mktime  1970 1 1 10 33 56      00 02 14   Or here s another way to do it with GNU date     date -ud      date -ud 1970-01-01 10 36 10    s -  date -ud 1970-01-01 10 33 56    s      T 00 02 14

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With GNU units     units 2411 units  71 prefixes  33 nonlinear units You have   10hr 36min 10s - 10hr 33min 56s  You want  s       134       0 0074626866 You have   10hr 36min 10s - 10hr 33min 56s  You want  min       2 2333333       0 44776119

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bin bash  START TIME   date   s   sleep 4  echo  Total time elapsed    date -ud        date   s  -  START TIME      T   HH MM SS         total time elapsed sh  Total time elapsed  00 00 04  HH MM SS

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Generalizing  nisetama s solution using GNU date  trusty Ubuntu 14 04 LTS    start  date     lt processing code gt  stop  date  duration  date -ud      date -ud  stop    s -  date -ud  start    s      T   echo  start echo  stop echo  duration   yielding   Wed Feb 7 12 31 16 CST 2018 Wed Feb 7 12 32 25 CST 2018 00 01 09

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Here is a solution using only the date commands capabilities using  ago   and not using a second variable to store the finish time      bin bash    save the current time start time    date   s  N      tested program sleep 1    the current time after the program has finished   minus the time when we started  in seconds nanoseconds elapsed time    date   s  N --date   start time seconds ago     echo elapsed time   elapsed time   this gives       time elapsed sh  elapsed time  1 002257120

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I needed a time difference script for use with mencoder  its --endpos is relative   and my solution is to call a Python script       timediff py 1 10 15 2 12 44 1 02 29   fractions of seconds are also supported     echo  diff is    timediff py 10 51 6 12 44   in hh mm ss format   diff is 0 01 52 4  in hh mm ss format    and it can tell you that the difference between 200 and 120 is 1h 20m       timediff py 120 0 200 0 1 20 0   and can convert any  probably fractional  number of seconds or minutes or hours to hh mm ss      timediff py 0 3600 1 00 0     timediff py 0 3 25 0 0 3 15 0   timediff py      usr bin python  import sys  def x60 h m       return 60 float h  float m   def seconds time       try         h m s   time split             return x60 x60 h m  s      except ValueError         try            m s   time split                return x60 m s         except ValueError            return float time   def difftime start  end       d   seconds end  - seconds start      print   d  02d  s     d 3600 d 60 60    02f     d 60   rstrip  0   rstrip        if   name         main        difftime sys argv 1  sys argv 2

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Seconds  To measure elapsed time  in seconds  we need    an integer that represents the count of elapsed seconds and a way to convert such integer to an usable format      An integer value of elapsed seconds    There are two bash internal ways to find an integer value for the number of elapsed seconds    Bash variable SECONDS  if SECONDS is unset it loses its special property     Setting the value of SECONDS to 0   SECONDS 0 sleep 1    Process to execute elapsedseconds  SECONDS  Storing the value of the variable SECONDS at the start   a  SECONDS sleep 1    Process to execute elapsedseconds     SECONDS - a      Bash printf option   datefmt T   a    TZ UTC0 printf     s T n   -1            -1   is the current time sleep 1                                      Process to execute elapsedseconds       TZ UTC0 printf     s T n   -1   - a          Convert such integer to an usable format  The bash internal printf can do that directly     TZ UTC0 printf     H  M  S T n  12345 03 25 45   similarly    elapsedseconds    12 60 34     TZ UTC0 printf     H  M  S T n    elapsedseconds  00 12 34   but this will fail for durations of more than 24 hours  as we actually print a wallclock time  not really a duration     hours 30 mins 12 secs 24   elapsedseconds         hours 60   mins  60   secs      TZ UTC0 printf     H  M  S T n    elapsedseconds  06 12 24   For the lovers of detail  from bash-hackers org        FORMAT T outputs the date-time string resulting from using FORMAT   as a format string for strftime 3   The associated argument is the   number of seconds since Epoch  or -1  current time  or -2  shell   startup time   If no corresponding argument is supplied  the current   time is used as default    So you may want to just call textifyDuration  elpasedseconds where textifyDuration is yet another implementation of duration printing   textifyDuration        local duration  1    local shiff  duration    local secs    shiff   60     shiff    shiff   60       local mins    shiff   60     shiff    shiff   60       local hours  shiff    local splur  if    secs  -eq 1    then splur     else splur  s   fi    local mplur  if    mins  -eq 1    then mplur     else mplur  s   fi    local hplur  if    hours -eq 1    then hplur     else hplur  s   fi    if     hours -gt 0     then       txt   hours hour hplur   mins minute mplur   secs second splur     elif     mins -gt 0     then       txt   mins minute mplur   secs second splur     else       txt   secs second splur     fi    echo   txt  from  duration seconds         GNU date   To get formated time we should use an external tool  GNU date  in several ways to get up to almost a year length and including Nanoseconds   Math inside date   There is no need for external arithmetic  do it all in one step inside date   date -u -d  0  FinalDate seconds -  StartDate seconds     H  M  S    Yes  there is a 0 zero in the command string  It is needed   That s assuming you could change the date    T  command to a date    s  command so the values will be stored  printed  in seconds   Note that the command is limited to    Positive values of  StartDate and  FinalDate seconds  The value in  FinalDate is bigger  later in time  than  StartDate  Time difference smaller than 24 hours  You accept an output format with Hours  Minutes and Seconds  Very easy to change  It is acceptable to use -u UTC times  To avoid  DST  and local time corrections      If you must use the 10 33 56 string  well  just convert it to seconds  also  the word seconds could be abbreviated as sec   string1  10 33 56  string2  10 36 10  StartDate   date -u -d   string1     s   FinalDate   date -u -d   string2     s   date -u -d  0  FinalDate sec -  StartDate sec     H  M  S    Note that the seconds time conversion  as presented above  is relative to the start of  this  day  Today      The concept could be extended to nanoseconds  like this   string1  10 33 56 5400022  string2  10 36 10 8800056  StartDate   date -u -d   string1     s  N   FinalDate   date -u -d   string2     s  N   date -u -d  0  FinalDate sec -  StartDate sec     H  M  S  N      If is required to calculate longer  up to 364 days  time differences  we must use the start of  some  year as reference and the format value  j  the day number in the year    Similar to   string1   10 days 10 33 56 5400022  string2   35 days 10 36 10 8800056  StartDate   date -u -d  2000 1 1  string1     s  N   FinalDate   date -u -d  2000 1 1  string2     s  N   date -u -d  2000 1 1  FinalDate sec -  StartDate sec     j days  H  M  S  N   Output  026 days 00 02 14 340003400   Sadly  in this case  we need to manually subtract 1 ONE from the number of days  The date command view the first day of the year as 1  Not that difficult      a     date -u -d  2000 1 1  FinalDate sec -  StartDate sec     j days  H  M  S  N     a 0     10   a 0  -1    echo    a         The use of long number of seconds is valid and documented here  https   www gnu org software coreutils manual html node Examples-of-date html Examples-of-date    Busybox date  A tool used in smaller devices  a very small executable to install   Busybox   Either make a link to busybox called date     ln -s  bin busybox date   Use it then by calling this date  place it in a PATH included directory    Or make an alias like     alias date  busybox date      Busybox date has a nice option  -D to receive the format of the input time  That opens up a lot of formats to be used as time  Using the -D option we can convert the time 10 33 56 directly   date -D   H  M  S  -d  10 33 56     Y  m  d- H  M  S    And as you can see from the output of the Command above  the day is assumed to be  today   To get the time starting on epoch     string1  10 33 56    date -u -D   Y  m  d- H  M  S  -d  1970 01 01- string1     Y  m  d- H  M  S  1970 01 01-10 33 56   Busybox date can even receive the time  in the format above  without -D     date -u -d  1970 01 01- string1     Y  m  d- H  M  S  1970 01 01-10 33 56   And the output format could even be seconds since epoch     date -u -d  1970 01 01- string1     s  52436   For both times  and a little bash math  busybox can not do the math  yet    string1  10 33 56  string2  10 36 10  t1   date -u -d  1970 01 01- string1     s   t2   date -u -d  1970 01 01- string2     s   echo     t2 - t1      Or formatted     date -u -D   s  -d      t2 - t1        H  M  S  00 02 14

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I realize this is an older post  but I came it across it today while working on a script that would take dates and times from a log file and compute the delta  The script below is certainly overkill  and I highly recommend checking my logic and maths      bin bash  dTime    tmp      firstEntry    head -n 1   LOG    sed  s        0-9  -         1     firstEntry  2013-01-16 01 56 37   lastEntry    tac   LOG    head -n 1   sed  s        0-9  -         1     lastEntry  2014-09-17 18 24 02     I like to make the variables easier to parse firstEntry    firstEntry  -     lastEntry    lastEntry  -     firstEntry    firstEntry        lastEntry    lastEntry           remove the following lines in production echo   lastEntry  echo   firstEntry     compute days in last entry for i in  seq 1   echo  lastEntry awk   print  2      do     case   i  in    1 3 5 7 8 10 12       dTime     dTime 31             4 6 9 11       dTime     dTime 30             2       dTime     dTime 28            esac   done    do leap year calculations for all years between first and last entry for i in  seq   echo  firstEntry awk   print  1      echo  lastEntry awk   print  1      do     if       i 4   -eq 0    amp  amp        i 100   -eq 0    amp  amp        i 400   -eq 0    then       if     i       echo  firstEntry awk   print  1        amp  amp      echo  firstEntry awk   print  2    -lt 2    then         dTime     dTime 1         elif     echo  firstEntry awk   print  2    -eq 2    amp  amp      echo  firstEntry awk   print  3    -lt 29    then         dTime     dTime 1         fi     elif       i 4   -eq 0    amp  amp        i 100   -ne 0    then       if     i       echo  lastEntry awk   print  1        amp  amp      echo  lastEntry awk   print  2    -gt 2    then         dTime     dTime 1         elif     echo  lastEntry awk   print  2    -eq 2    amp  amp      echo  lastEntry awk   print  3    -ne 29    then         dTime     dTime 1         fi     fi   done    substract days in first entry for i in  seq 1   echo  firstEntry awk   print  2      do     case   i  in    1 3 5 7 8 10 12       dTime     dTime-31             4 6 9 11       dTime     dTime-30             2       dTime     dTime-28            esac   done  dTime     dTime   echo  lastEntry awk   print  3   -  echo  firstEntry awk   print  3         The above gives number of days for sample  Now we need hours  minutes  and seconds   As a bit of hackery I just put the stuff in the best order for use in a for loop dTime       echo  lastEntry awk   print  6   -  echo  firstEntry awk   print  6           echo  lastEntry awk   print  5   -  echo  firstEntry awk   print  5           echo  lastEntry awk   print  4   -  echo  firstEntry awk   print  4       dTime  tmp 1 for i in  dTime  do     if    i -lt 0    then       case   tmp  in      1         tmp       echo  dTime awk   print  1    60        echo  dTime awk   print  2   -1          dTime   tmp   echo  dTime awk   print  3    4           tmp 1               2         tmp       echo  dTime awk   print  2    60        echo  dTime awk   print  3   -1          dTime    echo  dTime awk   print  1     tmp   echo  dTime awk   print  4           tmp 2               3         tmp       echo  dTime awk   print  3    24        echo  dTime awk   print  4   -1          dTime    echo  dTime awk   print  1    2     tmp        tmp 3              esac     fi   tmp     tmp 1     done  echo  The sample time is   echo  dTime awk   print  4    days    echo  dTime awk   print  3    hours    echo  dTime awk   print  2    minutes  and   echo  dTime awk   print  1    seconds     You will get output as follows   2012 10 16 01 56 37 2014 09 17 18 24 02 The sample time is 700 days  16 hours  27 minutes  and 25 seconds    I modified the script a bit to make it standalone  ie  just set variable values   but maybe the general idea comes across as well  You d might want some additional error checking for negative values

User · Answer

Here is how I did it   START   date   s   sleep 1    Your stuff END   date   s   echo    END-START     awk   print int  1 60    int  1 60      Really simple  take the number of seconds at the start  then take the number of seconds at the end  and print the difference in minutes seconds

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Here s some magic   time1 14 30 time2    date   H  M     16 00 diff     echo   time2 -  time1     sed  s     1 60   g    bc -l   echo  diff hours   outputs 1 5 hours   sed replaces a   with a formula to convert to 1 60  Then the time calculation that is made by bc

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I can t comment on mcaleaa s answer  hence I post this here  The  diff  variable should be on small case  Here is an example    root test     date1   date    s    date Wed Feb 21 23 00 20 MYT 2018  root test       root test     date2   date    s    date Wed Feb 21 23 00 33 MYT 2018  root test       root test     diff     date2- date1    root test        Previous variable was declared on lower case  This is what happened when upper case is used    root test     echo  Duration      DIFF   3600    hours      DIFF   3600    60   minutes     DIFF   60   seconds  -bash    3600   syntax error  operand expected  error token is    3600     root test        So  quick fix would be like this   root test     echo  Duration      diff   3600    hours      diff   3600    60   minutes     diff   60   seconds  Duration  0 hours 0 minutes 13 seconds  root test

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I d like to propose another way that avoid recalling date command  It may be helpful in case if you have already gathered timestamps in  T date format   ts get sec       read -r h m s  lt  lt  lt    echo  1   tr             echo     h 60 60   m 60  s      start ts 10 33 56 stop ts 10 36 10  START   ts get sec  start ts  STOP   ts get sec  stop ts  DIFF    STOP-START    echo     DIFF 60  m    DIFF 60  s    we can even handle millisecondes in the same way   ts get msec       read -r h m s ms  lt  lt  lt    echo  1   tr              echo     h 60 60 1000   m 60 1000   s 1000  ms      start ts 10 33 56 104 stop ts 10 36 10 102  START   ts get msec  start ts  STOP   ts get msec  stop ts  DIFF    STOP-START    min    DIFF  60 1000    sec     DIFF  60 1000   1000   ms     DIFF  60 1000   1000    echo    min    sec   ms

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As of date  GNU coreutils  7 4 you can now use -d to do arithmetic      date -d -30days Sat Jun 28 13 36 35 UTC 2014    date -d tomorrow Tue Jul 29 13 40 55 UTC 2014   The units you can use are days  years  months  hours  minutes  and seconds      date -d tomorrow 2days-10minutes Thu Jul 31 13 33 02 UTC 2014

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date can give you the difference and format it for you  OS X options shown   date -ujf s      date -jf T  10 36 10    s  -   date -jf T  10 33 56    s      T   00 02 14  date -ujf s      date -jf T  10 36 10    s  -   date -jf T  10 33 56    s             -Hh  -Mm  -Ss    0h 2m 14s   Some string processing can remove those empty values  date -ujf s      date -jf T  10 36 10    s  -   date -jf T  10 33 56    s             -Hh  -Mm  -Ss    sed  s     lt    0 hms     g    2m 14s   This won t work if you place the earlier time first  If you need to handle that  change      date      -   date        to   echo   date      -   date        bc   tr -d -

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Bash has a handy SECONDS builtin variable that tracks the number of seconds that have passed since the shell was started  This variable retains its properties when assigned to  and the value returned after the assignment is the number of seconds since the assignment plus the assigned value   Thus  you can just set SECONDS to 0 before starting the timed event  simply read SECONDS after the event  and do the time arithmetic before displaying   SECONDS 0   do some work duration  SECONDS echo      duration   60   minutes and     duration   60   seconds elapsed     As this solution doesn t depend on date   s  which is a GNU extension   it s portable to all systems supported by Bash

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start   date   s    echo  Diff    date -d       date   s - start      M minutes  S seconds    Diff  00 minutes 11 seconds

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Or wrap it up a bit  alias timerstart  starttime   date    s    alias timerstop  echo seconds      date    s  - starttime      Then this works   timerstart  sleep 2  timerstop seconds 2

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Here s my bash implementation  with bits taken from other SO  -   function countTimeDiff         timeA  1   09 59 35     timeB  2   17 32 55        feeding variables by using read and splitting with IFS     IFS   read ah am as  lt  lt  lt    timeA      IFS   read bh bm bs  lt  lt  lt    timeB         Convert hours to minutes        The 10  is there to avoid errors with leading zeros       by telling bash that we use base 10     secondsA    10  ah 60 60   10  am 60   10  as       secondsB    10  bh 60 60   10  bm 60   10  bs       DIFF SEC    secondsB - secondsA       echo  The difference is  DIFF SEC seconds         SEC     DIFF SEC 60       MIN      DIFF SEC- SEC  3600 60       HRS      DIFF SEC- MIN 60  3600       TIME DIFF   HRS  MIN  SEC       echo  TIME DIFF       countTimeDiff 2 15 55 2 55 16 The difference is 2361 seconds  0 39 21   Not tested  may be buggy

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Following on from Daniel Kamil Kozar s answer  to show hours minutes seconds  echo  quot Duration      DIFF   3600    hours      DIFF   3600    60   minutes     DIFF   60   seconds quot   So the full script would be  date1   date   quot  s quot   date2   date   quot  s quot   DIFF     date2- date1   echo  quot Duration      DIFF   3600    hours      DIFF   3600    60   minutes     DIFF   60   seconds quot

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Define this function  say in    bashrc    time  clock           -z   ts    amp  amp   ts  date   s N  return    te  date   s N      printf   6 4f    echo    te-ts   1000000000   bc -l      unset ts te     Now you can measure time of parts of your scripts     cat script sh       code     time  clock sleep 0 5 echo  Total time    time  clock         more code          script sh Total time  0 5060   very useful to find execution bottlenecks

[bash] How to calculate time elapsed in bash script?

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