CODEWITHSUNDEEP
bashcommand-line-arguments
ddd
ddd

Reputation: 5029

how to specify an command line argument with a variable in bash

I would like my bash script to take either 2 or 3 arguments with the last argument always being the input file to open. So the index of argument should depend on how many arguments provided. I know this can be realized by if statement, like:

if [ $# -eq 3 ]; then
  INFILE=$3
elif [ $# -eq 2 ]; then
  INFILE=$2
fi

..open file to read...

However I was hoping this to be done by a one liner which would look like this:

INFILE=$($#)

It does not work though. Same thing with INFILE=$"$#". Is it possible to specify index of argument directly with "$#"

Upvotes: 2

Views: 121

Answers (3)

Beggarman
Beggarman

Reputation: 896

You're trying to do indirect referencing of an array, so ${!#} would also work.

Upvotes: 1

ghoti
ghoti

Reputation: 46896

You're seeing a well known limitation in shell. :-)

You could get the last argument by stepping through the existing arguments:

#!/bin/bash

while [ ! -z "$2" ]; do
  shift
done

echo "Last argument: $1"

In Bourne (non-Bash) shell, you could do this with eval, which is evil:

#!/bin/bash

eval "echo \"Last argument: \$$#\""

If you don't need portability outside bash, you can also use the @ array:

#!/bin/bash

echo "Last argument: ${@: -1}"

Upvotes: 1

Rickkwa
Rickkwa

Reputation: 2291

Try using INFILE="${@: -1}" to get the last argument.

Upvotes: 3

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