Bash script not recognizing command-line arguments? [duplicate]

Question

I am calling the bash script mkproj.sh by using the command ./mkproj.sh. I also try calling it with arguments: ./mkproj.sh hello but my script returns a null $1 when I put echo "$1" in the script. I am not sure why it won't recognize the command-line arguments.

check_for_file()
{
if [ $# -eq 0 ]; then
        local testing_file=myproject
else
        local testing_file=$1
fi

if [ -d $testing_file ]; then
        echo "Directory name already exists"
        exit
else
        mkdir -p "$testing_file"/{archive,backups,docs/{html,txt},assets,database,src/{sh,c}}
fi

}
check_for_file

Thank you!

Please add a suitable shebang (#!/bin/bash) and then paste your script at shellcheck.net and try to implement the recommendations made there. — Cyrus
– Cyrus, Commented Oct 4, 2021 at 19:00
$1 is the first argument to the function. You called the function with no arguments, so $1 is not defined. — William Pursell
– William Pursell, Commented Oct 4, 2021 at 19:00

Arkadiusz Drabczyk · Accepted Answer · 2021-10-04 18:58:54Z

1

Positional parameters within functions shadow global positional parameters that script received. You need to call your function like that:

check_for_file "$1"

answered Oct 4, 2021 at 18:58

Arkadiusz Drabczyk

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Charles Duffy Over a year ago

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