Bash script to calculate time elapsed

Question

I am writing a script in bash to calculate the time elapsed for the execution of my commands  consider   STARTTIME   date   s   command block that takes time to complete              ENDTIME   date   s  echo  It takes    ENDTIME -  STARTTIME  seconds to complete this task       I guess my logic is correct however I end up with the following print out    It takes seconds to complete this task      Anything wrong with my string evaluation   I believe bash variables are untyped  I would love if there is a  string to integer  method in bash nevertheless

User · Answer

Try the following code   start   date    s    amp  amp  sleep 5  amp  amp  echo  It took      date    s   -  start   seconds

User · Answer

You are trying to execute the number in the ENDTIME as a command  You should also see an error like 1370306857  command not found  Instead use the arithmetic expansion   echo  It takes     ENDTIME -  STARTTIME   seconds to complete this task       You could also save the commands in a separate script  commands sh  and use time command   time commands sh

User · Answer

start   date   Y m d H M S   for x in  1  5   do echo  x  sleep 1  done  end   date   Y m d H M S   elapsed     end- start    ftime   for  i 1 i lt        end -   elapsed    i              do echo -n  -           done          echo   elapsed    echo -e  Start      start  nStop       end  nElapsed    ftime    Start    20171108005304 Stop     20171108005310 Elapsed  -------------6

User · Answer

I find it very clean to use the internal variable   SECONDS   SECONDS 0   sleep 10   echo  SECONDS

User · Answer

This is a one-liner alternative to Mike Q s function   secs to human         echo        1    3600   h        1    60    60   m       1    60   s

User · Answer

Either       or     will work for computing the result of an arithmetic operation   You re using     which is simply taking the string and evaluating it as a command   It s a bit of a subtle distinction   Hope this helps   As tink pointed out in the comments on this answer      is deprecated  and       should be favored

User · Answer

bin bash      time elapsed        appstop  1  appstart  2      ss strt   appstart 12 2   ss stop   appstop 12 2      mm strt   appstart 10 2   mm stop   appstop 10 2       hh strt   appstart 8 2    hh stop   appstop 8 2       dd strt   appstart 6 2    dd stop   appstop 6 2       mh strt   appstart 4 2    mh stop   appstop 4 2       yy strt   appstart 0 4    yy stop   appstop 0 4       if      ss stop   -lt    ss strt      then ss stop    ss stop 60    mm stop    mm stop-1    fi     if      mm stop   -lt  0     then mm stop    mm stop 60    hh stop    hh stop-1    fi     if      mm stop   -lt    mm strt      then mm stop    mm stop 60    hh stop    hh stop-1    fi     if      hh stop   -lt  0     then hh stop    hh stop 24    dd stop    dd stop-1    fi     if      hh stop   -lt    hh strt      then hh stop    hh stop 24    dd stop    dd stop-1    fi      if      dd stop   -lt  0     then dd stop    dd stop   mh days  mh stop  yy stop     mh stop    mh stop-1    fi     if      dd stop   -lt    dd strt      then dd stop    dd stop   mh days  mh stop  yy stop     mh stop    mh stop-1    fi      if      mh stop   -lt  0     then mh stop    mh stop 12    yy stop    yy stop-1    fi     if      mh stop   -lt    mh strt      then mh stop    mh stop 12    yy stop    yy stop-1    fi      ss espd    10   ss stop -10   ss strt     if       ss espd   -le  1     then ss espd   for  i 1 i lt        ss stop -   ss espd    i      do echo -n  0   done  echo   ss espd    fi     mm espd    10   mm stop -10   mm strt     if       mm espd   -le  1     then mm espd   for  i 1 i lt        mm stop -   mm espd    i      do echo -n  0   done  echo   mm espd    fi     hh espd    10   hh stop -10   hh strt     if       hh espd   -le  1     then hh espd   for  i 1 i lt        hh stop -   hh espd    i      do echo -n  0   done  echo   hh espd    fi     dd espd    10   dd stop -10   dd strt     if       dd espd   -le  1     then dd espd   for  i 1 i lt        dd stop -   dd espd    i      do echo -n  0   done  echo   dd espd    fi     mh espd    10   mh stop -10   mh strt     if       mh espd   -le  1     then mh espd   for  i 1 i lt        mh stop -   mh espd    i      do echo -n  0   done  echo   mh espd    fi     yy espd    10   yy stop -10   yy strt     if       yy espd   -le  1     then yy espd   for  i 1 i lt        yy stop -   yy espd    i      do echo -n  0   done  echo   yy espd    fi      echo -e    yy espd -  mh espd -  dd espd    hh espd    mm espd    ss espd        return   echo -e    yy espd -  mh espd -  dd espd    hh espd    mm espd    ss espd               mh days        mh stop  1  yy stop  2   also checks if it s leap year or not      case  mh stop in       1 3 5 7 8 10 12   mh stop 31              2       yy stop   4   amp  amp   yy stop   100      yy stop   400        amp  amp  mh stop 29    mh stop 28               4 6 9 11   mh stop 30             esac      return   mh stop             appstart   date   Y m d H M S   read -p  Wait some time  then press nay-key     key  appstop   date   Y m d H M S   elapsed   time elapsed  appstop  appstart   echo -e  Start       appstart 0 4 -  appstart 4 2 -  appstart 6 2    appstart 8 2    appstart 10 2    appstart 12 2  nStop        appstop 0 4 -  appstop 4 2 -  appstop 6 2    appstop 8 2    appstop 10 2    appstop 12 2  n  printf   0 1s      1  30   nElapsed     elapsed        exit 0   -------------------------------------------- return Wait some time  then press nay-key    Start     2017-11-09 03 22 17 Stop      2017-11-09 03 22 18                                Elapsed   0000-00-00 00 00 01

User · Answer

try using time with the elapsed seconds option    usr bin time -f e sleep 1 under bash   or  time -f e sleep 1 in interactive bash   see the time man page      Users  of  the  bash shell need to use an explicit path in order to run          the external time command and not the shell builtin variant  On  system          where time is installed in  usr bin  the first example would become                usr bin time wc  etc hosts   and   FORMATTING THE OUTPUT                A literal          e      Elapsed  real   wall  clock  time used by the process  in                  seconds

User · Answer

For larger numbers we may want to print in a more readable format  The example below does same as other but also prints in  human  format   secs to human         if    -z   1       1  -lt 60     then         min 0   secs    1       else         time mins   echo  scale 2    1  60    bc          min   echo   time mins    cut -d    -f1          secs  0   echo   time mins    cut -d    -f2           secs   echo   secs  60 bc awk   print int  1 0 5         fi     echo  Time Elapsed     min  minutes and   secs  seconds       Simple testing   secs to human  300  secs to human  305  secs to human  59  secs to human  60  secs to human  660  secs to human  3000    Output   Time Elapsed   5 minutes and 0 seconds  Time Elapsed   5 minutes and 5 seconds  Time Elapsed   0 minutes and 59 seconds  Time Elapsed   1 minutes and 0 seconds  Time Elapsed   11 minutes and 0 seconds  Time Elapsed   50 minutes and 0 seconds    To use in a script as described in other posts  capture start point then call the function with the finish time   start   date   s     lt  lt  performs some task here  gt  gt  secs to human       date   s  -   start

User · Answer

You can use Bash s time keyword here with an appropriate format string  TIMEFORMAT  It takes  R seconds to complete this task     time        command block that takes time to complete                       Here s what the reference says about TIMEFORMAT      The value of this parameter is used as a format string specifying how the timing information for pipelines prefixed with the time   reserved word should be displayed  The         character introduces an   escape sequence that is expanded to a time value or other information    The escape sequences and their meanings are as follows  the braces   denote optional portions           A literal            p  l R      The elapsed time in seconds    p  l U      The number of CPU seconds spent in user mode    p  l S      The number of CPU seconds spent in system mode   P      The CPU percentage  computed as   U    S     R         The optional p is a digit specifying the precision  the number of fractional digits after a decimal point  A value of 0 causes no   decimal point or fraction to be output  At most three places after the   decimal point may be specified  values of p greater than 3 are changed   to 3  If p is not specified  the value 3 is used       The optional l specifies a longer format  including minutes  of the form MMmSS FFs  The value of p determines whether or not the   fraction is included       If this variable is not set  Bash acts as if it had the value     nreal t 3lR nuser t 3lU nsys t 3lS        If the value is null  no timing information is displayed  A trailing newline is added when the format string is displayed

[linux] Bash script to calculate time elapsed

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