get end of day epoch in shell script bash

You can simply specify that with the -d argument.

i.e. date -d "today 23:59:59" +%swhere today 23:29:59 is used to get the end of the current day

Edit : @Toby propose the following approach to handle correctly leap seconds -d 'tomorrow 0 -1second

If you want the beginning of the day use

date -d "today 0" +%s

Using BSD date, you can use the -v to adjust the time properly.

% dateTue May 22 09:21:50 EDT 2018% date  -v 23H -v 59M -v 59STue May 22 23:59:59 EDT 2018

Slightly longer, but saving you from having to remember how many hours are in a day, minutes in an hour, etc, by going to midnight tomorrow, then subtracting one second.

% date -v +1d -v 0H -v 0M -v -0S -v -1S          ^      ^     ^      ^     ^          |      |     |      |     |          |      +-----+------+     subtract one second          |            |          |        reset to midnight          |          go to tomorrow

(It's a shame -v can't take a combination. date -v+1d0H0MS-1S would be nice to type. It's not terribly readable or obvious, but easy to parse if you know how -v works. Whitespace would make it better: date -v "+1d 0H 0M 0S -1S". #wishfulthinking)

A simple but fragile approach is to just round up to the nearest multiple of 86400:

$ now=$(date +%s)$ end_of_day=$(( now - now%86400 + 86399))

but this won't take daylight savings into account for the two days where it may be relevant.