Bash script: difference in minutes between two times

Question

I have two time strings; eg. \"09:11\" and \"17:22\" on the same day (format is hh:mm). How do I calculate the time difference in minutes between these two?

Can the

Answer 1

A pure bash solution :

old=09:11
new=17:22

# feeding variables by using read and splitting with IFS
IFS=: read old_hour old_min <<< "$old"
IFS=: read hour min <<< "$new"

# convert hours to minutes
# the 10# is there to avoid errors with leading zeros
# by telling bash that we use base 10
total_old_minutes=$((10#$old_hour*60 + 10#$old_min))
total_minutes=$((10#$hour*60 + 10#$min))

echo "the difference is $((total_minutes - total_old_minutes)) minutes"

Another solution using date (we work with hour/minutes, so the date is not important)

old=09:11
new=17:22

IFS=: read old_hour old_min <<< "$old"
IFS=: read hour min <<< "$new"

# convert the date "1970-01-01 hour:min:00" in seconds from Unix EPOCH time
sec_old=$(date -d "1970-01-01 $old_hour:$old_min:00" +%s)
sec_new=$(date -d "1970-01-01 $hour:$min:00" +%s)

echo "the difference is $(( (sec_new - sec_old) / 60)) minutes"

See http://en.wikipedia.org/wiki/Unix_time