How to do parsing of Elapsed time in seconds in linux - regex

Question

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.

Answer 1

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

Answer 2

The given value of A is*:
A="3day 3day/3h 15min/3h/15min 10sec/15min/10sec"
To compute A in seconds you can use bash's parameter expansion:
A=${A//day/*86400}
A=${A//h/*3600}
A=${A//min/*60}
A=${A//sec/*1}
A=${A//\//+}
A=${A// /+}
echo "A = $A"
echo $A | bc
Output:
A = 3*86400+3*86400+3*3600+15*60+3*3600+15*60+10*1+15*60+10*1
542720
* Note here I changed the original value of A as provided by the OP. From
3 day/3 day/3h...
to
3day 3day/3h... # the rest is the same as OP's.

Answer 3

Using awk to s/h/hours/ and to launch date +"%s" -d "1970-01-01 GMT +" to parse the time strings and to count the seconds:
$ awk '{
sub(/h/,"hours") # date no eat h
$1="" # remove $1
"date +\"%s\" -d \"1970-01-01 GMT + " $0 "\"" | getline s # date
print s
}' file
324000
259200
11700
10800
910
900
10
for the data:
$ cat file
1) 3 day 18h
2) 3 day
3) 3h 15min
4) 3h
5) 15min 10sec
6) 15min
7) 10sec