How do I forward parameters to other command in bash script?

Inside my bash script, I would like to parse zero, one or two parameters (the script can recognize them), then forward the remaining parameters to a command invoked in the script. How can I do that?

Use the shift built-in command to "eat" the arguments. Then call the child process and pass it the "$@" argument to include all remaining arguments. Notice the quotes, they should be kept, since they cause the expansion of the argument list to be properly quoted.

bash uses the shift command:

e.g. shifttest.sh:

#!/bin/bash
echo $1
shift
echo $1 $2

shifttest.sh 1 2 3 produces

1
2 3

Bash supports subsetting parameters (see Subsets and substrings), so you can choose which parameters to process/pass like this.

1. open new file and edit it: vim r.sh:

   echo "params only 2    : ${@:2:1}"
   echo "params 2 and 3   : ${@:2:2}"
   echo "params all from 2: ${@:2:99}"
   echo "params all from 2: ${@:2}"
   

2. run it:

   $ chmod u+x r.sh
   $ ./r.sh 1 2 3 4 5 6 7 8 9 10
   

3. the result is:

   params only 2    : 2
   params 2 and 3   : 2 3
   params all from 2: 2 3 4 5 6 7 8 9 10
   params all from 2: 2 3 4 5 6 7 8 9 10