How can I pass a file argument to my bash script using a Terminal command in Linux?

Question

So my question is how can I pass a file argument to my bash script using a Terminal command in Linux? At the moment I'm trying to make a program in bash that can take a file argument from the Terminal and use it as a variable in my program. For example I run myprogram --file=/path/to/file in Terminal.

My Program

#!/bin/bash    
File=(the path from the argument)  
externalprogram $File (other parameters)

How can I achieve this with my program?

Answer 1

It'll be easier (and more "proper", see below) if you just run your script as

myprogram /path/to/file

Then you can access the path within the script as $1 (for argument #1, similarly $2 is argument #2, etc.)

file="$1"
externalprogram "$file" [other parameters]

Or just

externalprogram "$1" [otherparameters]

If you want to extract the path from something like --file=/path/to/file, that's usually done with the getopts shell function. But that's more complicated than just referencing $1, and besides, switches like --file= are intended to be optional. I'm guessing your script requires a file name to be provided, so it doesn't make sense to pass it in an option.

Answer 2

you can use getopt to handle parameters in your bash script. there are not many explanations for getopt out there. here is an example:

#!/bin/sh

OPTIONS=$(getopt -o hf:gb -l help,file:,foo,bar -- "$@")

if [ $? -ne 0 ]; then
  echo "getopt error"
  exit 1
fi

eval set -- $OPTIONS

while true; do
  case "$1" in
    -h|--help) HELP=1 ;;
    -f|--file) FILE="$2" ; shift ;;
    -g|--foo)  FOO=1 ;;
    -b|--bar)  BAR=1 ;;
    --)        shift ; break ;;
    *)         echo "unknown option: $1" ; exit 1 ;;
  esac
  shift
done

if [ $# -ne 0 ]; then
  echo "unknown option(s): $@"
  exit 1
fi

echo "help: $HELP"
echo "file: $FILE"
echo "foo: $FOO"
echo "bar: $BAR"

see also:

• the "canonical" example: http://software.frodo.looijaard.name/getopt/docs/getopt-parse.bash
• a blog post: http://www.missiondata.com/blog/system-administration/17/17/
• man getopt

links dead. here from internet archive:

• https://web.archive.org/web/20100325015445/http://software.frodo.looijaard.name/getopt/docs/getopt-parse.bash

• https://web.archive.org/web/20130827004301/http://www.missiondata.com/blog/system-administration/17/17/

Answer 3

Bash supports a concept called "Positional Parameters". These positional parameters represent arguments that are specified on the command line when a Bash script is invoked.

Positional parameters are referred to by the names $0, $1, $2 ... and so on. $0 is the name of the script itself, $1 is the first argument to the script, $2 the second, etc. $* represents all of the positional parameters, except for $0 (i.e. starting with $1).

An example:

#!/bin/bash
FILE="$1"
externalprogram "$FILE" <other-parameters>

Answer 4

Assuming you do as David Zaslavsky suggests, so that the first argument simply is the program to run (no option-parsing required), you're dealing with the question of how to pass arguments 2 and on to your external program. Here's a convenient way:

#!/bin/bash
ext_program="$1"
shift
"$ext_program" "$@"

The shift will remove the first argument, renaming the rest ($2 becomes $1, and so on).$@` refers to the arguments, as an array of words (it must be quoted!).

If you must have your --file syntax (for example, if there's a default program to run, so the user doesn't necessarily have to supply one), just replace ext_program="$1" with whatever parsing of $1 you need to do, perhaps using getopt or getopts.

If you want to roll your own, for just the one specific case, you could do something like this:

if [ "$#" -gt 0 -a "${1:0:6}" == "--file" ]; then
    ext_program="${1:7}"
else
    ext_program="default program"
fi