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Linh Chi Nguyen
Asked: 2019-02-02 11:30:20 +0800 CST

how to assign a function with arguments to a variable, then pass the variable to another function in shell script

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I'm new to shell script, so any information is not trivial at all. I'm writing a generic function create_dir that take a variable $dirname and an option variable $function to generate a directory. If the optional $function is absent, the default function is mkdir. Like this:

File ~/bin/lib/create_dir.sh:

#!/bin/bash

create_dir()
{
    DIRNAME=$1
    FUNCTION=${2:-mkdir}
    $FUNCTION $DIRNAME
}
export -f create_dir

That works fine.

Now I import this into another file ~/bin/create_app

#!/bin/bash
. "${HOME}/bin/lib/create_dir.sh"

DIRNAME=$1
FUNCTION="python manage.py startapp"

create_dir $FUNCTION $DIRNAME

When I run create_app it imported the create_dir but the variable $FUNCTION it feeds the create_dir is wrong. The variable $FUNCTION has only one word python instead of python manage.py startapp as I wanted.

Why? How to fix it?

scripts command-line bash

1 Answers

  1. You should quote the argument like this: "$FUNCTION". Otherwise, the shell will run 

    create_dir python manage.py startapp ...

    followed by all the words in the first argument to create_app. This way the first argument to create_dir will be python and the second one manage.py.

    You are always on the safe side if you put double quotes around your variable expansions. This will prevent the shell from splitting the value of that variable into separate words, which the function then regards as different arguments when you pass to it that variable as an argument.

    There are cases where you can go without quotes, for example in the line

    DIRNAME=$1

    or more generally

    var2=$var

    as the bash will automatically assume that you want to assign the entire value of var to the variable var2. On the other hand, if you write the value directly, you need to use quotes:

    string="Hello World"
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