Timeline for Getting the last argument passed to a shell script

7 events

when toggle format	what		by	license	comment
Jun 8, 2020 at 11:06	comment	added	AgileZebra		Thanks dkasak. Updated to reflect your simplification.
Jun 8, 2020 at 11:05	history	edited	AgileZebra	CC BY-SA 4.0	deleted 7 characters in body
Jun 6, 2020 at 7:03	history	edited	oguz ismail	CC BY-SA 4.0	added 28 characters in body
Nov 4, 2016 at 20:02	comment	added	dkasak		AgileZebra's answer supplies a way of getting all but the last arguments so I wouldn't say Steven's answer supersedes it. However, there seems to be no reason to use `$((...))` to subtract the 1, you can simply use `${@:1:$# - 1}`.
Nov 26, 2015 at 8:05	comment	added	oHo		The Steven Penny's answer is a bit nicer: use `${@: -1}` for last and `${@: -2:1}` for second last (and so on...). Example: `bash -c 'echo ${@: -1}' prog 0 1 2 3 4 5 6` prints `6`. To stay with this current AgileZebra's approach, use `${@:$#-1:1}` to get the second last. Example: `bash -c 'echo ${@:$#-1:1}' prog 0 1 2 3 4 5 6` prints `5`. (and `${@:$#-2:1}` to get the third last and so on...)
Feb 24, 2015 at 11:51	history	edited	AgileZebra	CC BY-SA 3.0	changed per @starfry. Quite right, $# is all that'
Mar 31, 2011 at 6:30	history	answered	AgileZebra	CC BY-SA 2.5