How do I use $* while omitting certain input variables like $1 and $2 in bash script?

Question

For example,

elif [[ $append = $1 ]]
then
  touch ~/directory/"$2".txt
  echo "$variable_in_question" >> ~/directory/"$2".txt

to either create a text file containing all input following "$2" , or append an existing text file with all input following "$2", what would I use in place of the "$variable_in_question" in line 4?

I basically want "$*", but omitting "$1" and "$2".

Answer 1

You can use bash Parameter Expansion to specify a range, this works with positional parameters as well. For $3…$n it would be:

"${@:3}" # expands to "$3" "$4" "$5" …
"${*:3}" # expands to "$3 $4 $5 …"

Be aware that both $@ and $* ignore the first argument $0. If you wonder which one to use in your case: it’s very probable that you want a quoted $@. Don’t use $* unless you explicitly don’t want the arguments to be quoted individually.

You can try it out as follows:

$ bash -c 'echo "${@:3}"' 0 1 2 3 4 5 6
3 4 5 6
$ echo 'echo "${@:3}"' >script_file
$ bash script_file 0 1 2 3 4 5 6
2 3 4 5 6

Note that in the first example $0 is filled with the first argument 0 while when used in a script $0 is instead filled with the script’s name, as the second example shows. The script’s name to bash of course is the first argument, just that it's normally not perceived as such – the same goes for a script made executable and called “directly”. So in the first example we have $0=0, $1=1 etc. while in the second it’s $0=script_file, $1=0, $2=1 etc.; ${@:3} selects every argument starting with $3.

Some additional examples for possible ranges:

 # two arguments starting with the third
$ bash -c 'echo "${@:3:2}"' 0 1 2 3 4 5 6
3 4
 # every argument starting with the second to last one
 # a negative value needs either a preceding space or parentheses
$ bash -c 'echo "${@: -2}"' 0 1 2 3 4 5 6
5 6
 # two arguments starting with the fifth to last one
$ bash -c 'echo "${@:(-5):2}"' 0 1 2 3 4 5 6
2 3

Further reading:

• man bash/EXPANSION/Parameter Expansion
• bash-hackers.org: Handling positional parameters
• TLDP Advanced Bash-Scripting Guide: Parameter Substitution

Answer 2

You can use the shift builtin:

$ help shift
shift: shift [n]
    Shift positional parameters.

    Rename the positional parameters $N+1,$N+2 ... to $1,$2 ...  If N is
    not given, it is assumed to be 1.

    Exit Status:
    Returns success unless N is negative or greater than $#.

Ex. given

$ cat argtest.bash 
#!/bin/bash

shift 2

echo "$*"

then

$ ./argtest.bash foo bar baz bam boo
baz bam boo

Answer 3

Generally, you can copy the positional parameters to an array, delete arbitrary indices of the array and then use the array to expand to exactly those indices you want, without losing your original arguments.

For example, if I wanted all the args except the first, fourth and fifth:

args=( "$@" )
unset args[0] args[3] args[4]
echo "${args[@]}"

In the copy, the indices are shifted by 1, since $0 is not part of $@.