Mastering Time Arithmetic in Shell: Add, Subtract, and Compute Differences

Overview

Shell scripting often requires precise handling of dates and times, such as adding or subtracting intervals and calculating the difference between two timestamps. The following sections demonstrate reliable methods using date and Bash arithmetic.

1. Time Addition and Subtraction

The process consists of three steps: convert the original date to a Unix timestamp, convert the interval to seconds, and add the values together.

a. Convert the base time to a timestamp

time1=$(date +%s -d '1990-01-01 01:01:01')

Result:

631126861

b. Convert the interval to seconds

# 1 hour 20 minutes = 1*60*60 + 20*60 seconds
time2=$((1*60*60 + 20*60))
# Verify
echo $time2

Result:

4800

c. Add the timestamps and convert back to a readable date

time1=$(($time1 + $time2))
# Convert Unix timestamp to formatted date
date -d "1970-01-01 UTC $time1 seconds" '+%Y-%m-%d %H:%M:%S'

Result:

1990-01-01 02:21:01

2. Time Difference Calculation

To find the interval between two dates, convert each to a timestamp and subtract.

# Example: difference between 2010-01-01 and 2009-01-01 11:11:11
time1=$(($(date +%s -d '2010-01-01') - $(date +%s -d '2009-01-01 11:11:11')))
# Output the difference in seconds
echo $time1
# Convert to minutes if needed
echo $((time1 / 60))

The subtraction yields the total seconds; further division can produce minutes, hours, or days as required.

3. Bash Arithmetic Operators

Shell arithmetic can be performed with single or double parentheses. Single parentheses evaluate a command substitution, while double parentheses compute arithmetic expressions.

# Single parentheses (command substitution)
a=$(date)   # equivalent to a=date

# Double parentheses (arithmetic)
a=$((1+2))
echo $a    # prints 3

# Using expr as an alternative
a=$(expr 1 + 2)

These operators are essential for manipulating numeric values such as timestamps.