Remove first element from $@ in bash

Question

I'm writing a bash script that needs to loop over the arguments passed into the script. However, the first argument shouldn't be looped over, and instead needs to be checked before the loop.

If I didn't have to remove that first element I could just do:

for item in "$@" ; do
  #process item
done

I could modify the loop to check if it's in its first iteration and change the behavior, but that seems way too hackish. There's got to be a simple way to extract the first argument out and then loop over the rest, but I wasn't able to find it.

score 159 · Answer 1 · answered Apr 23 '10 at 20:44

159

Another variation uses array slicing:

for item in "${@:2}"
do
    process "$item"
done

This might be useful if, for some reason, you wanted to leave the arguments in place since shift is destructive.

answered Apr 23 '10 at 20:44

Dennis Williamson

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Thanks for ${@:2}. Good stuff! – cosmixtew Mar 05 '14 at 22:11
Exactly what I wanted. Now don't need temp variable for dereferencing first argument in `"${!1}${@:2}"` – Charlie Gorichanaz Mar 27 '17 at 22:59
@Herms this should be the accepted answer, more readable and not destructive (vs shift) – user1075613 Apr 03 '17 at 20:17
Beautiful, thanks a lot. – Jabari Dash Feb 08 '18 at 03:33
1

Technically the currently accepted answer is correct, but this is technically as well as practically correct (since shift is destructive). This is a better answer. So Ideally should be the answer. – Kuldeep Dhaka Aug 04 '18 at 16:32
For those who wonder, why `shift` is destructive: It actually removes the element ultimately from the arguments list. Ref: https://unix.stackexchange.com/a/174568/254204 – pico_prob Apr 30 '21 at 21:28

score 158 · Accepted Answer · edited Mar 01 '22 at 03:28

158

Use shift.

Read $1 for the first argument before the loop (or $0 if what you're wanting to check is the script name), then use shift, then loop over the remaining $@.

edited Mar 01 '22 at 03:28

Mateen Ulhaq

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answered Apr 23 '10 at 19:29

Amber

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See, I *knew* there was a simple answer. :) – Herms Apr 23 '10 at 19:36
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Deserves an upvote, but _after_ Dennis' answers get accepted :) – mgarciaisaia Apr 09 '14 at 03:47
@mgarciaisaia seems more appropriate to the OP's requirements: *remove* the first item – Mark Fox Oct 07 '14 at 02:49
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this is totally vague- not as helpful as @nos answer below – Charney Kaye Aug 30 '16 at 20:18
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Would be useful to have an example to avoid having to go to an external link that might disappear one day. – Gewthen Dec 17 '20 at 03:37
Link now points to a site flagged as a security risk by FireFox – Rob Osborne Feb 07 '21 at 23:30

score 57 · Answer 3 · edited Nov 15 '16 at 20:48

57

firstitem=$1
shift;
for item in "$@" ; do
  #process item
done

edited Nov 15 '16 at 20:48

twasbrillig

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answered Apr 23 '10 at 19:29

nos

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Remember that `$0` is generally the script name. – Amber Apr 23 '10 at 19:31
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+1 for showing a simple example; note that the `in "$@"` part is implied and can be omitted. – mklement0 Apr 09 '14 at 04:52
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@mklement0 You're saying you can just do "for item; do" ? – David Doria Apr 21 '14 at 11:57
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@DavidDoria: Yes, you can. Try `set -- 1 'two' 'three four'; for item; do echo "[$item]"; done` – mklement0 Apr 21 '14 at 13:29

James Andino · Answer 4 · 2014-04-09T04:38:05.933

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q=${@:0:1};[ ${2} ] && set ${@:2} || set ""; echo $q

EDIT

> q=${@:1}
# gives the first element of the special parameter array ${@}; but ${@} is unusual in that it contains (? file name or something ) and you must use an offset of 1;

> [ ${2} ] 
# checks that ${2} exists ; again ${@} offset by 1
    > && 
    # are elements left in        ${@}
      > set ${@:2}
      # sets parameter value to   ${@} offset by 1
    > ||
    #or are not elements left in  ${@}
      > set ""; 
      # sets parameter value to nothing

> echo $q
# contains the popped element

An Example of pop with regular array

   LIST=( one two three )
    ELEMENT=( ${LIST[@]:0:1} );LIST=( "${LIST[@]:1}" ) 
    echo $ELEMENT

edited Apr 09 '14 at 04:38

answered Apr 09 '14 at 03:37

James Andino

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Please also explain the code to be more educative. – lpapp Apr 09 '14 at 04:11
`q=${@:0:1}` (btw, you explanation misquotes it as `q=${@:1}`) should be `q=${@:1:1}` to be clearer : `$@` starts with index *1*, presumably to parallel the explicit `$1`, `$2`, ... parameters - element 0 has _no_ value (unlike its explicit counterpart, `$0`, which reflects the shell / script file). `[ ${2} ]` will _break_ should `$2` contain embedded spaces; you won't have that problem if you use `[[ ${2} ]]` instead. That said, the conditional and the `||` branch are not needed: if there is only 1 argument, `${@:2}` will simply expand to the empty string (cont'd). – mklement0 Apr 09 '14 at 05:05
(cont'd) You should, however, use `set --` so as to ensure that arguments that happen to look like options are not interpreted as such by `set` and you should double-quote the `${@:2}` reference and `$q` reference in the `echo` statement. At this point we get: `q=${@:1:1}; set -- "${@:2}"; echo "$q"`. However, `q=${@:1:1}` is just another (more cumbersome) way of saying `$1`, and the rest essentially re-implements the `shift` command; using these features we simply get: `q=$1; shift; echo "$q"`. – mklement0 Apr 09 '14 at 05:08
@mklement0 I was wonder about where a good place to place a printf to clean the paths ? ( I wish I could be clearer ) – James Andino Apr 09 '14 at 06:26
Unfortunately, I don't understand. What paths? Clean in what way? – mklement0 Apr 09 '14 at 13:38

Remove first element from $@ in bash

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