Script parameters in Bash

Question

I'm trying to make a shell script which should be used like this:

ocrscript.sh -from /home/kristoffer/test.png -to /home/kristoffer/test.txt

The script will then ocr convert the image file to a text file. Here is what I have come up with so far:

#!/bin/bash
export HOME=/home/kristoffer
/usr/local/bin/abbyyocr9 -rl Swedish -if ???fromvalue??? -of ???tovalue??? 2>&1

But I don't know how to get the -from and -to values. Any ideas on how to do it?

Answer 1

The arguments that you provide to a bashscript will appear in the variables $1 and $2 and $3 where the number refers to the argument. $0 is the command itself.

The arguments are seperated by spaces, so if you would provide the -from and -to in the command, they will end up in these variables too, so for this:

./ocrscript.sh -from /home/kristoffer/test.png -to /home/kristoffer/test.txt

You'll get:

$0    # ocrscript.sh
$1    # -from
$2    # /home/kristoffer/test.png
$3    # -to
$4    # /home/kristoffer/test.txt

It might be easier to omit the -from and the -to, like:

ocrscript.sh /home/kristoffer/test.png /home/kristoffer/test.txt

Then you'll have:

$1    # /home/kristoffer/test.png
$2    # /home/kristoffer/test.txt

The downside is that you'll have to supply it in the right order. There are libraries that can make it easier to parse named arguments on the command line, but usually for simple shell scripts you should just use the easy way, if it's no problem.

Then you can do:

/usr/local/bin/abbyyocr9 -rl Swedish -if "$1" -of "$2" 2>&1

The double quotes around the $1 and the $2 are not always necessary but are adviced, because some strings won't work if you don't put them between double quotes.