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How to do parsing of Elapsed time in seconds in linux

I want to do parsing of Elapsed time in seconds .Time formats given below:

1) 3 day 18h
2) 3 day
3) 3h 15min
4) 3h 
5) 15min 10sec
6) 15min 
7) 10sec

i'm getting values from systemctl status cassandra | awk '/(Active: active)/{print $9, $10,$11}' Now storing it's value in variable A,like

A=$(systemctl status cassandra | awk '/(Active: active)/{print $9, $10,$11}'

now A has input as 3 day 18h or 3 day etc. More examples-

A=3 day 18h or 3 day or 3h 15min or 3h or 15min 10sec or 15min or 10sec

now take different values of A, and parse in seconds.

Solution

  • What you want to achieve could be done directly in awk using the following line :

    $ systemctl status cassandra | awk '/(Active: active)/{s=$6" "$7;gsub(/-|:/," ",s); print systime() - mktime(s)}'

    This will give you the running time directly based on the start-time and not on the approximated running time printed by systemctl.

    If this approach is not working then I suggest to use the date command to do all the parsing. If you can change the h by hour in your examples, then you can do the following :

    $ date -d "1970-01-01 + 3day 18hour 15min 16sec" +%s
324916

    If you cannot, then I suggest the following. If duration is stored in the variable $duration, then you do

    $ date -d "1970-01-01 + ${duration/h/hour}" +%s

    Having spaces between the numbers and the strings day, h,min or sec does not matter.

    The idea of this is that you ask date to compute everything for you as %s returns the unix time since 1970-01-01 in seconds.

    man date: %s seconds since 1970-01-01 00:00:00 UTC