How best to include other scripts? - bash

Question

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?

Answer 1

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"

Answer 2

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 an 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

Answer 3

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)

Answer 4

If it is in the same directory you can use dirname $0:
#!/bin/bash
source $(dirname $0)/incl.sh
echo "The main script"

Answer 5

I think the best way to do this is to use the Chris Boran's way, BUT you should compute MY_DIR this way:
#!/bin/sh
MY_DIR=$(dirname $(readlink -f $0))
$MY_DIR/other_script.sh
To quote the man pages for readlink:
readlink - display value of a symbolic link
...
-f, --canonicalize
canonicalize by following every symlink in every component of the given
name recursively; all but the last component must exist
I've never encountered a use case where MY_DIR is not correctly computed. If you access your script through a symlink in your $PATH it works.

Answer 6

A combination of the answers to this question provides the most robust solution.
It worked for us in production-grade scripts with great support of dependencies and directory structure:
#!/bin/bash
# Full path of the current script
THIS=`readlink -f "${BASH_SOURCE[0]}" 2>/dev/null||echo $0`
# The directory where current script resides
DIR=`dirname "${THIS}"`
# 'Dot' means 'source', i.e. 'include':
. "$DIR/compile.sh"
The method supports all of these:
Spaces in path
Links (via readlink)
${BASH_SOURCE[0]} is more robust than $0

Answer 7

SRC=$(cd $(dirname "$0"); pwd)
source "${SRC}/incl.sh"

Answer 8

1. Neatest
I explored almost every suggestion and here is the neatest one that worked for me:
script_root=$(dirname $(readlink -f $0))
It works even when the script is symlinked to a $PATH directory.
See it in action here: https://github.com/pendashteh/hcagent/blob/master/bin/hcagent
2. The coolest
# Copyright https://stackoverflow.com/a/13222994/257479
script_root=$(ls -l /proc/$$/fd | grep "255 ->" | sed -e 's/^.\+-> //')
This is actually from another answer on this very page, but I'm adding it to my answer too!
3. The most reliable
Alternatively, in the rare case that those didn't work, here is the bullet proof approach:
# Copyright http://stackoverflow.com/a/7400673/257479
myreadlink() { [ ! -h "$1" ] && echo "$1" || (local link="$(expr "$(command ls -ld -- "$1")" : '.*-> \(.*\)$')"; cd $(dirname $1); myreadlink "$link" | sed "s|^\([^/].*\)\$|$(dirname $1)/\1|"); }
whereis() { echo $1 | sed "s|^\([^/].*/.*\)|$(pwd)/\1|;s|^\([^/]*\)$|$(which -- $1)|;s|^$|$1|"; }
whereis_realpath() { local SCRIPT_PATH=$(whereis $1); myreadlink ${SCRIPT_PATH} | sed "s|^\([^/].*\)\$|$(dirname ${SCRIPT_PATH})/\1|"; }
script_root=$(dirname $(whereis_realpath "$0"))
You can see it in action in taskrunner source: https://github.com/pendashteh/taskrunner/blob/master/bin/taskrunner
Hope this help someone out there :)
Also, please leave it as a comment if one did not work for you and mention your operating system and emulator. Thanks!

Answer 9

This works even if the script is sourced:
source "$( dirname "${BASH_SOURCE[0]}" )/incl.sh"

Answer 10

You need to specify the location of the other scripts, there is no other way around it. I'd recommend a configurable variable at the top of your script:
#!/bin/bash
installpath=/where/your/scripts/are
. $installpath/incl.sh
echo "The main script"
Alternatively, you can insist that the user maintain an environment variable indicating where your program home is at, like PROG_HOME or somesuch. This can be supplied for the user automatically by creating a script with that information in /etc/profile.d/, which will be sourced every time a user logs in.

Answer 11

I'd suggest that you create a setenv script whose sole purpose is to provide locations for various components across your system.
All other scripts would then source this script so that all locations are common across all scripts using the setenv script.
This is very useful when running cronjobs. You get a minimal environment when running cron, but if you make all cron scripts first include the setenv script then you are able to control and synchronise the environment that you want the cronjobs to execute in.
We used such a technique on our build monkey that was used for continuous integration across a project of about 2,000 kSLOC.

Answer 12

Shell Script Loader is my solution for this.
It provides a function named include() that can be called many times in many scripts to refer a single script but will only load the script once. The function can accept complete paths or partial paths (script is searched in a search path). A similar function named load() is also provided that will load the scripts unconditionally.
It works for bash, ksh, pd ksh and zsh with optimized scripts for each one of them; and other shells that are generically compatible with the original sh like ash, dash, heirloom sh, etc., through a universal script that automatically optimizes its functions depending on the features the shell can provide.
[Fowarded example]
start.sh
This is an optional starter script. Placing the startup methods here is just a convenience and can be placed in the main script instead. This script is also not needed if the scripts are to be compiled.
#!/bin/sh
# load loader.sh
. loader.sh
# include directories to search path
loader_addpath /usr/lib/sh deps source
# load main script
load main.sh
main.sh
include a.sh
include b.sh
echo '---- main.sh ----'
# remove loader from shellspace since
# we no longer need it
loader_finish
# main procedures go from here
# ...
a.sh
include main.sh
include a.sh
include b.sh
echo '---- a.sh ----'
b.sh
include main.sh
include a.sh
include b.sh
echo '---- b.sh ----'
output:
---- b.sh ----
---- a.sh ----
---- main.sh ----
What's best is scripts based on it may also be compiled to form a single script with the available compiler.
Here's a project that uses it: http://sourceforge.net/p/playshell/code/ci/master/tree/. It can run portably with or without compiling the scripts. Compiling to produce a single script can also happen, and is helpful during installation.
I also created a simpler prototype for any conservative party that may want to have a brief idea of how an implementation script works: https://sourceforge.net/p/loader/code/ci/base/tree/loader-include-prototype.bash. It's small and anyone can just include the code in their main script if they want to if their code is intended to run with Bash 4.0 or newer, and it also doesn't use eval.

Answer 13

Steve's reply is definitely the correct technique but it should be refactored so that your installpath variable is in a separate environment script where all such declarations are made.
Then all scripts source that script and should installpath change, you only need to change it in one location. Makes things more, er, futureproof. God I hate that word! (-:
BTW You should really refer to the variable using ${installpath} when using it in the way shown in your example:
. ${installpath}/incl.sh
If the braces are left out, some shells will try and expand the variable "installpath/incl.sh"!

Answer 14

I put all my startup scripts in a .bashrc.d directory.
This is a common technique in such places as /etc/profile.d, etc.
while read file; do source "${file}"; done <<HERE
$(find ${HOME}/.bashrc.d -type f)
HERE
The problem with the solution using globbing...
for file in ${HOME}/.bashrc.d/*.sh; do source ${file};done
...is you might have a file list which is "too long".
An approach like...
find ${HOME}/.bashrc.d -type f | while read file; do source ${file}; done
...runs but doesn't change the environment as desired.

Answer 15

This should work reliably:
source_relative() {
local dir="${BASH_SOURCE%/*}"
[[ -z "$dir" ]] && dir="$PWD"
source "$dir/$1"
}
source_relative incl.sh

Answer 16

Using source or $0 will not give you the real path of your script. You could use the process id of the script to retrieve its real path
ls -l /proc/$$/fd |
grep "255 ->" |
sed -e 's/^.\+-> //'
I am using this script and it has always served me well :)

Answer 17

Of course, to each their own, but I think the block below is pretty solid. I believe this involves the "best" way to find a directory, and the "best" way to call another bash script:
scriptdir=`dirname "$BASH_SOURCE"`
source $scriptdir/incl.sh
echo "The main script"
So this may be the "best" way to include other scripts. This is based off another "best" answer that tells a bash script where it is stored

Answer 18

Personally put all libraries in a lib folder and use an import function to load them.
folder structure
script.sh contents
# Imports '.sh' files from 'lib' directory
function import()
{
local file="./lib/$1.sh"
local error="\e[31mError: \e[0mCannot find \e[1m$1\e[0m library at: \e[2m$file\e[0m"
if [ -f "$file" ]; then
source "$file"
if [ -z $IMPORTED ]; then
echo -e $error
exit 1
fi
else
echo -e $error
exit 1
fi
}
Note that this import function should be at the beginning of your script and then you can easily import your libraries like this:
import "utils"
import "requirements"
Add a single line at the top of each library (i.e. utils.sh):
IMPORTED="$BASH_SOURCE"
Now you have access to functions inside utils.sh and requirements.sh from script.sh
TODO: Write a linker to build a single sh file

Answer 19

we just need to find out the folder where our incl.sh and main.sh is stored; just change your main.sh with this:
main.sh
#!/bin/bash
SCRIPT_NAME=$(basename $0)
SCRIPT_DIR="$(echo $0| sed "s/$SCRIPT_NAME//g")"
source $SCRIPT_DIR/incl.sh
echo "The main script"

Answer 20

According man hier suitable place for script includes is /usr/local/lib/
/usr/local/lib
Files associated with locally installed programs.
Personally I prefer /usr/local/lib/bash/includes for includes.
There is bash-helper lib for including libs in that way:
#!/bin/bash
. /usr/local/lib/bash/includes/bash-helpers.sh
include api-client || exit 1 # include shared functions
include mysql-status/query-builder || exit 1 # include script functions
# include script functions with status message
include mysql-status/process-checker; status 'process-checker' $? || exit 1
include mysql-status/nonexists; status 'nonexists' $? || exit 1

Answer 21

Most of the answers I saw here seem to overcomplicate things. This method has always worked reliably for me:
FULLPATH=$(readlink -f $0)
INCPATH=${FULLPATH%/*}
INCPATH will hold the complete path of the script excluding the script filename, regardless of how the script is called (by $PATH, relative or absolute).
After that, one only needs to do this to include files in the same directory:
. $INCPATH/file_to_include.sh
Reference: TecPorto / Location independent includes

Answer 22

here is a nice function you can use. it builds on what #sacii made. thank you
it will let you list any number of space separated script names to source (relative to the script calling source_files).
optionally you can pass an absolute or relative path as the first argument and it will source from there instead.
you can call it multiple times (see example below) to source scripts from different dirs
#!/usr/bin/env bash
function source_files() {
local scripts_dir
scripts_dir="$1"
if [ -d "$scripts_dir" ]; then
shift
else
scripts_dir="${BASH_SOURCE%/*}"
if [[ ! -d "$scripts_dir" ]]; then scripts_dir="$PWD"; fi
fi
for script_name in "$#"; do
# shellcheck disable=SC1091 disable=SC1090
. "$scripts_dir/$script_name.sh"
done
}
here is an example you can run to show how its used
#!/usr/bin/env bash
function source_files() {
local scripts_dir
scripts_dir="$1"
if [ -d "$scripts_dir" ]; then
shift
else
scripts_dir="${BASH_SOURCE%/*}"
if [[ ! -d "$scripts_dir" ]]; then scripts_dir="$PWD"; fi
fi
for script_name in "$#"; do
# shellcheck disable=SC1091 disable=SC1090
. "$scripts_dir/$script_name.sh"
done
}
## -- EXAMPLE -- ##
# assumes dir structure:
# /
# source_files.sh
# sibling.sh
# scripts/
# child.sh
# nested/
# scripts/
# grandchild.sh
cd /tmp || exit 1
# sibling.sh
tee sibling.sh <<- EOF > /dev/null
#!/usr/bin/env bash
export SIBLING_VAR='sibling var value'
EOF
# scripts/child.sh
mkdir -p scripts
tee scripts/child.sh <<- EOF > /dev/null
#!/usr/bin/env bash
export CHILD_VAR='child var value'
EOF
# nested/scripts/grandchild.sh
mkdir -p nested/scripts
tee nested/scripts/grandchild.sh <<- EOF > /dev/null
#!/usr/bin/env bash
export GRANDCHILD_VAR='grandchild var value'
EOF
source_files 'sibling'
source_files 'scripts' 'child'
source_files 'nested/scripts' 'grandchild'
echo "$SIBLING_VAR"
echo "$CHILD_VAR"
echo "$GRANDCHILD_VAR"
rm sibling.sh
rm -rf scripts nested
cd - || exit 1
prints:
sibling var value
child var value
grandchild var value

Answer 23

You can also use:
PWD=$(pwd)
source "$PWD/inc.sh"