How to do parsing of Elapsed time in seconds in linux

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