Can I call a function of a shell script from another shell script?

I have 2 shell scripts.

The second shell script contains following functions second.sh

func1 
func2

The first.sh will call the second shell script with some parameters and will call func1 and func2 with some other parameters specific to that function.

Here is the example of what I am talking about

second.sh

val1=`echo $1`
val2=`echo $2`

function func1 {

fun=`echo $1`
book=`echo $2`

}

function func2 {

fun2=`echo $1`
book2=`echo $2`


}

first.sh

second.sh cricket football

func1 love horror
func2 ball mystery

How can I achieve it?

Refactor your second.sh script like this:

func1 {
   fun="$1"
   book="$2"
   printf "func=%s,book=%s\n" "$fun" "$book"
}

func2 {
   fun2="$1"
   book2="$2"
   printf "func2=%s,book2=%s\n" "$fun2" "$book2"
}

And then call these functions from script first.sh like this:

source ./second.sh
func1 love horror
func2 ball mystery

OUTPUT:

func=love,book=horror
func2=ball,book2=mystery

You can't directly call a function in another shell script.

You can move your function definitions into a separate file and then load them into your script using the . command, like this:

. /path/to/functions.sh

This will interpret functions.sh as if it's content were actually present in your file at this point. This is a common mechanism for implementing shared libraries of shell functions.

The problem

The currenly accepted answer works only under important condition. Given...

/foo/bar/first.sh:

function func1 {  
   echo "Hello $1"
}

and

/foo/bar/second.sh:

#!/bin/bash

source ./first.sh
func1 World

this works only if the first.sh is executed from within the same directory where the first.sh is located. Ie. if the current working path of shell is /foo, the attempt to run command

cd /foo
./bar/second.sh

prints error:

/foo/bar/second.sh: line 4: func1: command not found

That's because the source ./first.sh is relative to current working path, not the path of the script. Hence one solution might be to utilize subshell and run

(cd /foo/bar; ./second.sh)

More generic solution

Given...

/foo/bar/first.sh:

function func1 {  
   echo "Hello $1"
}

and

/foo/bar/second.sh:

#!/bin/bash

source $(dirname "$0")/first.sh

func1 World

then

cd /foo
./bar/second.sh

prints

Hello World

How it works

• $0 returns relative or absolute path to the executed script
• dirname returns relative path to directory, where the $0 script exists
• $( dirname "$0" ) the dirname "$0" command returns relative path to directory of executed script, which is then used as argument for source command
• in "second.sh", /first.sh just appends the name of imported shell script
• source loads content of specified file into current shell

second.sh

#!/bin/bash

function func1() {
   fun="$1"
   book="$2"
   echo "$fun, $book\n"
}

function func2() {
   fun2="$1"
   book2="$2"
   printf "$fun2, $book2\n"
}

first.sh

#!/bin/bash

source /absolute_path_to/second.sh
func1 love horror
func2 ball mystery

You need to keep these things in your mind before using it

• Keyword source and . (a period) are the same command.
• If the FILENAME is not a full path to a file, the command will search for the file in the directories specified in the $PATH environmental variable . If the file is not found in the $PATH, the command will look for the file in the current directory.
• If any ARGUMENTS are given, they will become positional parameters to the FILENAME.
• If the FILENAME exists, the source command exit code is 0, otherwise, if the file is not found it will return 1.

Among these points the point to be focused on is the second one, you actually need to provide a ABSOLUTE_PATH to the file if you are using #!/bin/bash, RELATIVE_PATH just doesn't work if that is the case with you then my friend you just need to change the path to ABSOLUTE_FILE_PATH.