How to include file in a bash shell script

Question

Is there a way to include another shell script in a shell script to be able to access its functions?

Like how in PHP you can use the include directive with

Answer 1

Simply put inside your script :

source FILE

Or

. FILE # POSIX compliant

$ LANG=C help source
source: source filename [arguments]
Execute commands from a file in the current shell.

Read and execute commands from FILENAME in the current shell.  The
entries in $PATH are used to find the directory containing FILENAME.
If any ARGUMENTS are supplied, they become the positional parameters
when FILENAME is executed.

Exit Status:
Returns the status of the last command executed in FILENAME; fails if
FILENAME cannot be read.

Answer 2

Syntax is source <file-name>

ex. source config.sh

script - config.sh

USERNAME="satish"
EMAIL="satish@linuxconcept.com"

calling script -

#!/bin/bash
source config.sh
echo Welcome ${USERNAME}!
echo Your email is ${EMAIL}.

You can learn to include a bash script in another bash script here.

Answer 3

Yes, use source or the short form which is just .:

. other_script.sh

Answer 4

In my situation, in order to include color.sh from the same directory in init.sh, I had to do something as follows.

. ./color.sh

Not sure why the ./ and not color.sh directly. The content of color.sh is as follows.

RED=`tput setaf 1`
GREEN=`tput setaf 2`
BLUE=`tput setaf 4`
BOLD=`tput bold`
RESET=`tput sgr0`

Making use of File color.sh does not error but, the color do not display. I have tested this in Ubuntu 18.04 and the Bash version is:

GNU bash, version 4.4.19(1)-release (x86_64-pc-linux-gnu)

Answer 5

Above answers are correct, but if run script in other folder, there will be some problem.

For example, the a.sh and b.sh are in same folder, a include b with . ./b.sh to include.

When run script out of the folder, for example with xx/xx/xx/a.sh, file b.sh will not found: ./b.sh: No such file or directory.

I use

. $(dirname "$0")/b.sh