Bash: How

The way you would normally include a script is with "source"

eg:

main.sh:

#!/bin/bash

source incl.sh

echo "The main script"

incl.sh:

echo "The included script"

The output of executing "./main.sh" is:

The included script
The main script

... Now, if you attempt to execute that shell script from another location, it can't find the include unless it's in your path.

What's a good way to ensure that your script can find the include script, especially if for instance, the script needs to be portable?

---

I tend to make my scripts all be relative to one another. That way I can use dirname:

#!/bin/sh

my_dir="$(dirname "$0")"

"$my_dir/other_script.sh"

---

I know I am late to the party, but this should work no matter how you start the script and uses builtins exclusively:

DIR="${BASH_SOURCE%/*}"
if [[ ! -d "$DIR" ]]; then DIR="$PWD"; fi
. "$DIR/incl.sh"
. "$DIR/main.sh"

. (dot) command is alias to source, $PWD is the Path for the Working Directory, BASH_SOURCE is an array variable whose members are the source filenames, ${string%substring} strips shortest match of $substring from back of $string

---

An alternative to:

scriptPath=$(dirname $0)

is:

scriptPath=${0%/*}

.. the advantage being not having the dependence on dirname, which is not a built-in command (and not always available in emulators)

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