Date subtraction in bash - bash

Question

I want to get successive differences of dates shown sample below. How do I convert each of these dates into epoch seconds?
7/21/17 6:39:12:167 GMT
7/21/17 6:39:12:168 GMT
7/21/17 6:39:12:168 GMT
7/21/17 6:39:12:205 GMT
7/21/17 6:39:12:206 GMT
7/21/17 6:39:12:206 GMT
Once each line gets converted into epoch seconds, I can simply run another script to get successive differentials of each. Thanks.

Answer 1

You can convert times using the date command. Given a line like this:
7/21/17 6:39:12:167 GMT
You first need to strip everything at and after the seconds part, to get this:
7/21/17 6:39:12
You can use cut -d: -f1-3 for that. Then, convert to epoch seconds, if you're using FreeBSD or Mac OS:
date -ujf "%m/%d/%y %H:%M:%S" "7/21/17 6:39:12" +%s
Which gives:
1500619152
If you are using GNU date (e.g. on Linux), you can feed an entire file of dates to it. Since the input file isn't in the right format, we can do this:
date --file <(cut -d: -f1-3 infile) +%s
That will read the entire file with only a single invocation of date, which is much more efficient, but only works with GNU date.

Answer 2

Here is one in GNU awk. It converts the timestamps to epoch time in seconds and subtracts the former from the latter. mktime function used for converting doesn't take fraction of a seconds but the fractions are stored to hash a[7] and nothing stops you from adding it to t var before subtracting:
$ awk '
function zeropad(s) { # zeropad function
return sprintf("%02d", s)
}
{
split($0,a,"[/ :]") # split timestamp to a
for(i in a)
a[i]=zeropad(a[i]) # zeropad all components
t=mktime(20 a[3] " " a[1] " " a[3] " " a[4] " " a[5] " " a[6])
# add the fractions in a[7] here
if(NR>1) # might be unnecessary
print t-p # output subtracted seconds
p=t # set current time to previous
}' file
0
0
0
0
0
Since you didn't include the expected output or proper data sample, that's the best I can do for now.
EDIT:
Since your data does not fully reflect if the fraction of a second are presented like 0:0:0:100 or 0:0:0:1 I modified the zeropad function to left and right pad given values. Now you call it like zeropad(value, count, left/right) or zeropad(a[7],3,"r"):
function zeropad(s,c,d) {
return sprintf("%" (d=="l"? "0" c:"") "d" (d=="r"?"%0" c-length(s) "d":""), s,"")
}
{
split($0,a,"[/ :]") # split timestamp to a
for(i in a)
a[i]=zeropad(a[i],2,"l") # left-pad all components with 0s
t=mktime(20 a[3] " " a[1] " " a[3] " " a[4] " " a[5] " " a[6])
t=t+zeropad(a[7],3,"r")/1000 # right-pad fractions with 0s
if(NR>1) # might be unnecessary
print t-p # output subtracted seconds
p=t # set current time to previous
}
0.00999999
0
0.37
0.00999999
0
printf with proper modifiers should probably be used for outputing to get sane values.

Answer 3

Let's suppose that you have one of those dates in a variable:
$ d='7/21/17 6:39:12:167 GMT'
With GNU date, we need to remove the milliseconds part. That can be done with bash's pattern substitution:
$ echo "${d/:??? / }"
7/21/17 6:39:12 GMT
To convert that to seconds-since-epoch, we use the -d option to set the date and the %s format to request seconds-since-epoch:
$ date -d "${d/:??? / }" +%s
1500619152
Compatibility: Mac OSX does not support GNU's -d option but I gather it has similar option. On many operating systems, including OSX, there is the option to install GNU utilities.