Time difference in seconds between given two dates - bash

Question

I have two dates as follows:
2019-01-06 00:02:10 | END
2019-01-05 23:52:00 | START
How could I calculate and print the difference between START and END dates in seconds?
For above case I would like to get something like:
610

Answer 1

Assuming GNU implementation based OS, you can use date's option %s and -d to calculate the time difference in seconds using command substitution and arithmetic operations.
START="2019-01-05 23:52:00"
END="2019-01-06 00:02:10"
Time_diff_in_secs=$(($(date -d "$END" +%s) - $(date -d "$START" +%s)))
echo $Time_diff_in_secs
Output:
610
Hope this helps!!!

Answer 2

With bash and GNU date:
while read d t x x; do
[[ $x == "END" ]] && end="$d $t"
[[ $x == "START" ]] && start="$d $t"
done < file
end=$(date -u -d "$end" '+%s')
start=$(date -u -d "$start" '+%s')
diff=$(($end-$start))
echo "$diff"
Output:
610
See: man date

Answer 3

What you're asking for is difficult verging on impossible using pure bash. Bash doesn't have any date functions of its own. For date processing, most recommendations you'll get will be to use your operating system's date command, but the usage of this command varies by operating system.
In BSD (including macOS):
start="2019-01-05 23:52:00"; end="2019-01-06 00:02:10"
printf '%d\n' $(( $(date -j -f '%F %T' "$end" '+%s') - $(date -j -f '%F %T' "$start" '+%s') ))
In Linux, or anything using GNU date (possibly also Cygwin):
printf '%d\n' $(( $(date -d "$end" '+%s') - $(date -d "$start" '+%s') ))
And just for the fun of it, if you can't (or would prefer not to) use date for some reason, you might be able to get away with gawk:
gawk 'END{ print mktime(gensub(/[^0-9]/," ","g",end)) - mktime(gensub(/[^0-9]/," ","g",start)) }' start="$start" end="$end" /dev/null
The mktime() option parses a date string in almost exactly the format you're providing, making the math easy.

Answer 4

START="2019-01-05 23:52:00"
END="2019-01-06 00:02:10"
parse () {
local data=(`grep -oP '\d+' <<< "$1"`)
local y=$((${data[0]}*12*30*24*60*60))
local m=$((${data[1]}*30*24*60*60))
local d=$((${data[2]}*24*60*60))
local h=$((${data[3]}*60*60))
local mm=$((${data[4]}*60))
echo $((y+m+d+h+mm+${data[5]}))
}
START=$(parse "$START")
END=$(parse "$END")
echo $((END-START)) // OUTPUT: 610

Answer 5

Was trying to solve the same problem on a non-GNU OS, i.e. macOS. I couldn't apply any of the solutions above, although it inspired me to come up with the following solution. I am using some in-line Ruby from within my shell script, which should work out of the box on macOS.
START="2019-01-05 23:52:00"
END="2019-01-06 00:02:10"
SECONDS=$(ruby << RUBY
require 'date'
puts ((DateTime.parse('${END}') - DateTime.parse('${START}')) * 60 * 60 * 24).to_i
RUBY)
echo ${SECONDS}
# 610