LinuxQuestions.org - How to pass arguments in a shell script?

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How to pass arguments in a shell script?

Hello,
i am getting use to use shell scripts for linux.
How do I pass arguments?
Basically I found a shell where I can compare files and append the ones not found:

if ! diff file1 file2 > /dev/null; then
echo diff
else
echo not_diff
fi

But I want to replace file1 and file2 as two arguments...how can I do this? Thanks

Hi,

$1, $2, $3 etc represent the arguments give to the script at run time.

somescript.sh foo bar foobar

Inseide the somescript.sh script foo is assigned to $1, bar to $2, foobar to $3.

Although you can work with $1, $2 etc inside your script it is wise to use variables that are human readable.

Code:

#!/bin/bash



inFileOne="$1"

inFileTwo="$2"



if ! diff ${inFileOne} ${inFileTwo} > /dev/null; then

  <some code>

else

  <some other code>

fi

The above can be started as: scriptname.sh file1 file2

Do take a look at the Bash guides that are around. Here are 2:
Bash Guide for Beginners
Advanced Bash-Scripting Guide

Hope this helps.