Using the value of a variable in the name of an argument

Question

I'm writing a bash script which will input arguments, the command would look like this:

command -a -b -c file -d -e

I would like to detect a specific argument (-b) with its specific location ($1, $2, $3)

#! /bin/bash
counter=0
while [ counter -lt $# ]
do
    if [ $($counter) == "-b" ]
    then
        found=$counter
    fi
    let counter+=1
done

The problem rises in $($counter). Is there a way to use $counter to call the value of an argument? For instance if counter=2, I would like to call the value of argument $2. $($counter) doesn't work.

Use `getopts` for this. For example http://stackoverflow.com/a/14203146/1983854 can help. — fedorqui, Oct 16 '13 at 22:54
ok thanks! I'll look into it, but if I want the following filename, for instance -b filename, does `getopts` also provide this function? — Michael Ho Chum, Oct 16 '13 at 22:57
Yes it does. Did you look at the [documentation](http://linux.die.net/man/1/getopt)? — Carl Norum, Oct 16 '13 at 22:57

chepner · Accepted Answer · 2013-10-17T13:52:31.637

2

You can accomplish this without getopts (which is still recommended, though) by reworking your loop.

counter=1
for i in "$@"; do
  if [[ $i == -b ]]; then
      break
  fi
  ((counter+=1))
done

Simply iterate over the arguments directly, rather than iterating over the argument positions.

bash also does allow indirect parameter expansion, using the following syntax:

#! /bin/bash
counter=0
while [ counter -lt $# ]
do
    if [ ${!counter} = "-b" ]  # ${!x} uses the value of x as the parameter name
    then
        found=$counter
    fi
    let counter+=1
done

edited Oct 17 '13 at 13:52

answered Oct 17 '13 at 13:31

chepner

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Don't use `==` with single-bracket notation, `bash` accepts it but it's not POSIX-conforming and other shells choke on it. +1 for the rest! – Adrian Frühwirth Oct 17 '13 at 13:38
Non-POSIX shells will probably choke on `${!counter}` anyway :) But you are correct in general. – chepner Oct 17 '13 at 13:52

Using the value of a variable in the name of an argument

1 Answers1