Hello - I wish to count the number of day’s between two dates.

This can be achieved by using -

$ date
Tue Feb 23 13:24:42 GMT 2016

echo $(( ($(date -d 'Fri Feb 05 09:55:37 GMT 2016' +%s) - $(date +%s) )/ 86400)) days

Result -

-18 days

But I'm seeking a result of 19 days, where the current partial day counts as 1 full day, producing 19 days.

Any ideas?

Since you're discarding the fractions of days, subtract 86399 seconds from it before you calculate your day count. This will round it into the next day.

I tried this hack but it still fails on the second instance - where it give 1 day, instead the result sought is 2 days.

$ date
Tue Feb 23 13:24:42 GMT 2016

$ mydate=$(( ($(date +%s) - $(date -d 'Mon Feb 22 09:55:37 GMT 2016' +%s) )/ 86400)) && mydate=$((mydate+1)) && printf "%s days", $mydate
2 days

$ mydate=$(( ($(date +%s) - $(date -d 'Mon Feb 22 19:55:37 GMT 2016' +%s) )/ 86400)) && mydate=$((mydate+1)) && printf "%s days", $mydate
1 days

So I did some further testing -

$ date
Tue Feb 23 14:08:12 GMT 2016

This example works -

$ echo $(( ($(date -d 'Fri Feb 22 13:55:37 GMT 2016' +%s) - $(date +%s) - 86399)/ 86400)) days
-2 days

But this fails - the result sought here is still two days, but 1 day is produced.

$ echo $(( ($(date -d 'Fri Feb 22 15:55:37 GMT 2016' +%s) - $(date +%s) - 86399)/ 86400)) days
-1 days

What is being sought - 1 day = round up each day, even if only 1 second past midnight.

You should type "set -xv" in your terminal window so it can print out some debug to show you what all the calculations are doing.

Debug is not giving sufficient insight into the inner workings -

$ echo $(( ($(date -d 'Fri Feb 22 15:55:37 GMT 2016' +%s) - $(date +%s) - 86399)/ 86400)) days
echo $(( ($(date -d 'Fri Feb 22 15:55:37 GMT 2016' +%s) - $(date +%s) - 86399)/ 86400)) days
date -d 'Fri Feb 22 15:55:37 GMT 2016' +%s
++ date -d 'Fri Feb 22 15:55:37 GMT 2016' +%s
date +%s
++ date +%s
+ echo -1 days
-1 days

Write it in a script. Save each term value into a variable. Echo out each variable and each resultant calculation. That will help you to resolve your border cases.

What I'm saying is that I'm not convinced that the 86399 value doesn't work. Prove to me that it doesn't.

Consider that you are comparing exactly 1 day and one second's difference and you now add the 86399 value to that which is one second less than an entire day. The calculation result then becomes 2 days, which was your intentions. If you are exactly 1 day and you add 86399 you end up at one second under 2 days and the calculation result should then result in 1 day, which was your stated intentions.

The relative time difference between Feb 22 15:55:37 GMT 2016 and Feb 23 14:08:12 GMT 2016 is less than 24 hours (-0.925) so rounding down to -1 day is correct. What rtmistler posted works but not working as expected.

Not exactly sure what you are trying to accomplish but it isn't necessary to specify time when trying to find the difference between dates. Without time the default is 00:00:00.

bash doesn't do floating point and rounds down. As michaelk said, your result is less than 24 hours, which causes it to round down. Store your variables and then pass them to bc for the math part if you want the full answer.

When I've been working with comparing dates, it seems using the epoch date (date +%s) is the only way to get what I've been looking for.

From the man page for date:

%s seconds since 1970-01-01 00:00:00 UTC

Example:

# date +%s
1456253803
#


Then we can do the math however we like.

The following has done the trick -

$ now=$(date +%s) && then=$(date -d 'Tue Feb 22 13:55:37 GMT 2016' +%s) && count=$(echo "($then - $now) / 86400" | bc -l) && echo $count |awk -F'-' '{printf "%0.f\n", $2}'
2

now=$(date +%s) && then=$(date -d 'Tue Feb 23 13:55:37 GMT 2016' +%s) && count=$(echo "($then - $now) / 86400" | bc -l) && echo $count |awk -F'-' '{printf "%0.f\n", $2}'
1

now=$(date +%s) && then=$(date -d 'Wed Feb 24 13:55:37 GMT 2016' +%s) && count=$(echo "($then - $now) / 86400" | bc -l) && echo $count |awk -F'-' '{printf "%0.f\n", $2}'
0

Would appear to be a perfect use-case for dateutils.

1 members found this post helpful.

Why are you subtracting it in the wrong direction and then using awk to remove the negative sign? Just subtract it in the right order.