Cannot open file when pass $1, but can open file when pass the file name

January 20, 2022

This might be a naive question, but I am new to bash scripting. I am trying to open a file and count the number of lines in that file and if it is greater than 4, then output something. I have figured out how to pass and save the line count to a variable, but I cannot access the file in the bash when I do not pass it the exact name. Therefore, I have this program called new_test written as so:

test_function()
{
    file_name='test_text.txt'
    n=$(wc -l < $file_name)
    if [ 'n > 2' ]
    then
        echo "Too many lines"
    fi
}
test_function

This program works, but when I try and make it more generic to accept any file by changing line 3 from:

file_name='test_text.txt'

file_name=$1

And then call

./new_test test_text.txt

it outputs:

./test_file[4]: : cannot open

From my understanding, the $1 is the second argument passed, and in the case of ./new_test test_text.txt, this would be the test_text.txt file. Do I need to obtain the address of the second argument or is the notation different in order to do this? Thank you in advance.

>Solution :

The $1 inside of the function is the function argument, not the script argument. You need to take that top level argument and pass it back into the function itself.

test_function()
{
    file_name="$1"
    n=$(wc -l < $file_name)
    if [ 'n > 2' ]
    then
        echo "Too many lines"
    fi
}
test_function "$1"

Make sure you don’t forget the quotes either, or you’ll run into problems with files that have spaces in their path.