mjhjmtu

Is it possible to use bash to subtract variables containing 24-hour time?

#!/bin/bash
var1="23:30" # 11:30pm
var2="20:00" # 08:00pm

echo "$(expr $var1 - $var2)"

Running it produces the following error.

./test 
expr: non-integer argument

I need the output to appear in decimal form, for example:

./test 
3.5

The date command is pretty flexible about its input. You can use that to your advantage:

#!/bin/bash
var1="23:30"
var2="20:00"

# Convert to epoch time and calculate difference.
difference=$(( $(date -d "$var1" "+%s") - $(date -d "$var2" "+%s") ))

# Divide the difference by 3600 to calculate hours.
echo "scale=2 ; $difference/3600" | bc

Output:

$ ./test.bash
3.50

Using only bash, with no external programs, you could do so something like this:

#!/bin/bash

# first time is the first argument, or 23:30 
var1=$1:-23:30
# second time is the second argument, or 20:00
var2=$2:-20:00

# Split variables on `:` and insert pieces into arrays
IFS=':' read -r -a t1 <<< "$var1"
IFS=':' read -r -a t2 <<< "$var2"

# strip leading zeros (so it's not interpreted as octal
t1=("$t1[@]##0")
t2=("$t2[@]##0")

# check if the first time is before the second one
if (( t1[0] > t2[0] || ( t1[0] == t2[0] && t1[1] > t2[1]) ))
then
 # if the minutes on the first time are less than the ones on the second time
 if (( t1[1] < t2[1] ))
 then
 # add 60 minutes to time 1
 (( t1[1] += 60 ))
 # and subtract an hour
 (( t1[0] -- ))
 fi
 # now subtract the hours and the minutes
 echo $((t1[0] -t2[0] )):$((t1[1] - t2[1]))
 # to get a decimal result, multiply the minutes by 100 and divide by 60
 echo $((t1[0] -t2[0] )).$(((t1[1] - t2[1])*100/60))
else
 echo "Time 1 should be after time 2" 2>&1
fi

Test:

$ ./script.sh 
3:30
3.50

$ ./script.sh 12:10 11:30
0:40
0.66

$ ./script.sh 12:00 11:30
0:30
0.50

If you want more complex time differences, that could span different days etc, then it's probably best to use GNU date.