[linux] Script parameters in Bash

In bash $1 is the first argument passed to the script, $2 second and so on

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

So you can use:

./your_script.sh some_source_file.png destination_file.txt

Explanation on double quotes;

consider three scripts:

# foo.sh
bash bar.sh $1

# cat foo2.sh
bash bar.sh "$1"

# bar.sh
echo "1-$1" "2-$2"

Now invoke:

$ bash foo.sh "a b"
1-a 2-b

$ bash foo2.sh "a b"
1-a b 2-

When you invoke foo.sh "a b" then it invokes bar.sh a b (two arguments), and with foo2.sh "a b" it invokes bar.sh "a b" (1 argument). Always have in mind how parameters are passed and expaned in bash, it will save you a lot of headache.

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