从bash中的$@中删除第一个元素[duplicate]

使用 shift ?

http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_09_07.html

基本上,读 $1 循环之前的第一个参数(或 $0 如果要检查的是脚本名,则使用 转移 ,然后循环剩余的 $@ .

另一个变体使用数组切片:

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

如果出于某种原因,您希望保留争论的位置 shift 是破坏性的。

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

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

编辑

> 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

具有正则数组的pop示例

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