Remove first element from $@ in bash [duplicate]

This question already has answers here: Process all arguments except the first one (in a bash script) (5 answers) Closed 7 years ago.

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 ch开发者_开发百科ecked 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.

---

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 $@.

---

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.

---

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

---

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