functions with their own parameters whitin a bash script

Question

I want to create a script on Linux to create some directories if they do not exists. The directories are declared inside the script itself (not being passed from command line):

#!/bin/bash

varBaseDir="/home/user"

# Directories to create
varAppDir="${varBaseDir}/app"
varAppDataDir="${varBaseDir}/appData"

if [[ -d ${varAppDir} ]]
        then
                echo "app dir exist"
        else
                echo "app dir does not exist! creating '${dir}' ..."
                mkdir -p ${dir}
                if test "$?" -eq "0"
                        then
                                        echo "succeeded."
                        else
                                echo "failed to create directory"
                fi
fi

As you can see I create one of the directories. Then I would need to repeat the same code again but with a different directory (next time I want the varAppDataDir to be created).

I tried to figure out how to declare a function:

my_function($dir) {
        echo $dir
}

my_funciotn("HELLO")

but I get error:

./test.sh: line 27: syntax error near unexpected token `$dir'
./test.sh: line 27: `my_function($dir) {'

So how to do this right?

Answer 1

Functions don't have named arguments in POSIX shell or Bash. Instead, the arguments show up in the positional parameters, $1, $2, etc.

You'll have to manually copy them to named local variables if you want that. So:

foo() {
    local this="$1"
    local that="$2"
    printf "foo: that is '%s'\n" "$that"
}
foo one two

(local variables aren't specified in the standard, but pretty much all shells except ksh (ksh93?) treat them the same.)

See:

• https://mywiki.wooledge.org/BashGuide/CompoundCommands#Functions
• https://www.gnu.org/software/bash/manual/html_node/Shell-Functions.html