Iterate through parameters skipping the first

Hi i have the following:

bash_script parm1 a b c d ..n

I 开发者_C百科want to iterate and print all the values in the command line starting from a, not from parm1

You can "slice" arrays in bash; instead of using shift, you might use

for i in "${@:2}"
do
    echo "$i"
done

$@ is an array of all the command line arguments, ${@:2} is the same array less the first element. The double-quotes ensure correct whitespace handling.

This should do it:

#ignore first parm1
shift

# iterate
while test ${#} -gt 0
do
  echo $1
  shift
done

This method will keep the first param, in case you want to use it later

#!/bin/bash

for ((i=2;i<=$#;i++))
do
  echo ${!i}
done

or

for i in ${*:2} #or use $@
do
  echo $i
done

Another flavor, a bit shorter that keeps the arguments list

shift
for i in "$@"
do
  echo $i
done

You can use an implicit iteration for the positional parameters:

shift
for arg
do
    something_with $arg
done

As you can see, you don't have to include "$@" in the for statement.