How can I subtract timestamp results from `date` in one line of shell? - bash

Question

I can generate two timestamps like so:
date +"%s" -d "$(curl -s --head http://google.com | grep ^Date: | sed 's/Date: //g')"
// Result: 1417800327
date +"%s"
// Result: 1417800325
How can I subtract them with only one line?
echo "$((1417800327-1417800325))"
// Result: 2
But I want something closer to:
echo "$(( (date +"%s" -d "$(curl -s --head http://google.com | grep ^Date: | sed 's/Date: //g')") - (date +"%s")))"

Answer 1

Try doing this :
LANG=C echo $(( $(date +%s) - $(date -d "$(curl -s --head http://google.com 2>&1 | awk -F'Date: ' '/^Date:/{print $2}')" +%s) ))
or splitted multi-lines for readability :
LANG=C echo $((
$(date +%s) - $(
date -d "$(curl -s --head http://google.com |
awk -F'Date: ' '/^Date:/{print $2}'
)" +%s)
))

Answer 2

#!/bin/bash
##On Linux. start time, do something, find end time.
st=`date +%s`; sleep 10; et=`date +%s`;
##On SunOS instead of using date command, use `truss /usr/bin/date 2>&1 | grep ^time | awk -F"= " '{print $2}'`;
##substract to find elapsed time
elt=$((et-st));
##find elapsed hours, minutes, seconds
eh=$((elt/3600)); em=$(($((elt%3600))/60)); es=$(($((elt%3600))%60));
elt=`printf "%02d:%02d:%02d%s" $((eh)) $((em)) $((es))`;
##show elapsed time in user friendly way
echo - Elapsed time: $elt
Output:
- Elapsed time: 00:00:10

Answer 3

Using a subshell function with one big command bash line
(timestamp=`date +"%s" -d "$(curl -s --head http://google.com | grep ^Date: | sed 's/Date: //g')"`;date=`date +"%s"`;let "time_difference= $timestamp-$date";echo "$time_difference"; )