bash: time schedule and loops script - bash

Question

I'm a bash enthusiast. I've written many scripts, but this time I think this one is too big for me. I'm working on a script that would check the number of hours based on the schedule from the file and sum them up.
I have a file with weekly schedule. By using grep it returns me time schedule (for example 13:00 - 20:30 or 13:30 - 21:30). Those files will come always in the same format, only the hours will be different so my grep query will always work. The only difference is that sometimes it will return four lines (with hours), sometimes less, sometimes more, dependes what's in the given file (I can always use grep -c if the number of those lines matters).
When I subtract the finishing time from the starting time, the result is correct, although it won't work if the hours are a bit odd (for example 14:00 - 16:17). How can I achive that (2h and 17minutes)?
And then I think I should use a loop. I tried a couple of things and got super confused and decided to come here.
Here's what I've got so far (26to31August is obviously the file with the schedule).
#!/bin/bash
#COUNT=$(grep -c '[0-9][0-9]:[0-9][0-9]' 26to31August)
FINISH=$(grep '[0-9][0-9]:[0-9][0-9]' 26to31August |head -1 |awk '{ print $3 }')
START=$(grep '[0-9][0-9]:[0-9][0-9]' 26to31August |head -1 |awk '{ print $1 }')
VAR1=$(date -d "${FINISH}" +%s)
VAR2=$(date -d "${START}" +%s)
HOURS=$((VAR1-VAR2))
SECS2HOURS=$(expr $HOURS / 3600)
echo "That day it's $SECS2HOURS hours"
I want the script to echo each of those lines telling me how many hours it is per line and then add them all together.
Should I use For loop? Should I use a function in the loop?
I'll apprecieate your answers.

Answer 1

This may not be what you want to hear, but this is definitely a case where a different programming language will work infinitely better than Bash arithmetic. In Python, for example, you would simply do duration = finish - start. That will do a timezone-aware, leap day-aware, sane calculation resulting in an object from which you can get everything from years to fractional seconds.

Answer 2

Yes you would use a loop. You should maintain the running total in seconds and convert to hours:minutes in the final output
First thought (warning, this code contains a bug)
total=0
grep '[0-9][0-9]:[0-9][0-9]' 26to31August | while read -r start x finish x; do
a=$( date -d "$start" "+%s" )
b=$( date -d "$finish" "+%s" )
(( total += b - a ))
done
# output
hours=$(( total / 3600 ))
minutes=$(( (total % 2600) / 60 ))
printf "%d:%02d\n" $hours $minutes
bash can parse a whitespace-separated string into words with the read command.
Where's the bug? With bash's default configuration, each command in a pipeline is executed in a separate subshell. Therefore, the total variable's contents will be lost when the subshell running the while loop ends.
Here are two ways to get around that:
group the output commands into the same subshell
total=0
grep '[0-9][0-9]:[0-9][0-9]' 26to31August | {
while read -r start x finish x; do
a=$( date -d "$start" "+%s" )
b=$( date -d "$finish" "+%s" )
(( total += b - a ))
done
hours=$(( total / 3600 ))
minutes=$(( (total % 2600) / 60 ))
printf "%d:%02d\n" $hours $minutes
}
don't use a pipe, redirect the output of a process substitution
into the while loop
total=0
while read -r start x finish x; do
a=$( date -d "$start" "+%s" )
b=$( date -d "$finish" "+%s" )
(( total += b - a ))
done < <( grep '[0-9][0-9]:[0-9][0-9]' 26to31August )
hours=$(( total / 3600 ))
minutes=$(( (total % 2600) / 60 ))
printf "%d:%02d\n" $hours $minutes
Note, don't use ALLCAPS variable names. It's too easy to accidentally overwrite crucial shell variables.