CODEWITHSUNDEEP
linuxbashshellunix
latech
latech

Reputation: 73

Trouble reading second argument in bash script

I need to pass argument for commit function. When I do the commit through

./test.sh commit -m "first"

its not really committing it. believe somehow I am not passing right argument parameter either in case or function.

Here is the script 

#!/usr/bin/env bash

clone () {
  git clone $1
}

commit () {
  git commit $*
}

case $1
in
   clone) clone $2 ;;
   commit) commit $2 ;;

       *) echo "Invalid Argument passed" ;;
esac

Upvotes: 0

Views: 1803

Answers (2)

pjh
pjh

Reputation: 8064

To safely support multiple arguments (including ones with special characters) the function bodies should be

git clone "$@"

and

git commit "$@"

.

For the same reasons, the case code should be:

case $1 in
    clone)  clone "${@:2}" ;;
    commit) commit "${@:2}" ;;
    *)      echo "Invalid Argument passed" ;;
esac

In the functions, "$@" expands to all the function arguments, safely quoted so they are not subject to word splitting or expansions.

In the case statement, ${@:2} expands to the list of command line arguments after the first one, safely quoted.

For more information see Handling positional parameters [Bash Hackers Wiki].

Upvotes: 1

Nic3500
Nic3500

Reputation: 8601

The arguments are processed like this by bash:

./test.sh commit -m "first" 

0: ./test.sh
1: commit
2: -m
3: first

So your "first" is actually argument $3.

Upvotes: 1

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