mjhjmtu

I have a script at the moment that reads in variables and then operates on them like the following;

#!bin/bash
a=10
b=15
c=20

d=a*b+c
echo $d

However I would like to split this up into an input file containing:

a=10
b=15
c=20

and a script which does the operation

#!/bin/bash
d=a*b+c
echo $d

And will be called up something like this.

./script.sh < input.in

Now I have done a bit of digging and I have tried doing a simple.

./script.sh < input.in

such that it returns the answer 170

but this doesn't work. After further digging, it seems that in the script in need to use the command "source" But I am unsure how to do it in this case.

Can it be done? What is the best way to do this?

From 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.

So, in the script, you only need to add this line:

source input.in

or this one (the POSIX version):

. input.in

To pass the input file at runtime, you use positional parameters:

source "$1"

. "$1"

Also note that d=a*b+c won't work unless d has the "integer" attribute:

declare -i d
d=a*b+c

or you use arithmetic expansion to perform the operation:

d=$((a*b+c))

Example:

#!/bin/bash
source "$1"
d=$((a*b+c))
echo "$d"

$ ./script.sh input.in
170

This is a very basic operation. Just source or . a file name, and the result is as if those lines were included in your primary script.

The syntax is simply this:

source file_name

or

. file_name

There are some nuances, but that's more advanced. See man bash | grep source for more details.