Question

I know php programming but I am not familiar with bash script programming syntax.

I have this code snippet

EXPECTEDARGS=0
if [ $# -ne $EXPECTEDARGS -o "x$0" == "x"  -o $0 == "bash" ]; then
  echo "Usage:"
  echo "   Parameter 1: argument 1 missing"
  exit 1
fi

This correctly checks for at least 1 argument.

I want to understand what this code snippet means line by line. So far, I can only make out that EXPECTEDARGS is a variable.

After understanding the code snippet, I want to modify it to check for 2 arguments.

The code snippet will be in a github gist.

I aim to execute the script using

bash -c "$(curl -fsSL https://raw.github.com/gist/4491019)" <arg1> <arg2> 

without the <> symbols of course.

Answer 1

• $# number of arguments.
• $0 shell script name
• $1,$2,…,$n argument 1, 2, etc
• [ condition ] test one or more conditions, -ne not equal, -eq equal, -gt greater than, -lt lesser than, -o or, = compare string/number
• “x$0” = x check if script name/prog name is empty

if you want to know more, save the script as a file and try to print stuff with echo. Run the script with:

bash -c ./script.sh

or

bash script.sh

to mimic the behaviour of bash+curl

[ ] or test is a good example, they are just a program, actually the same program, while checking $0 or argv[0] in C, they behave a little bit different.

$ ls -l /bin/test /bin/\[
-r-xr-xr-x  2 root  wheel  43120 Dec 11  2011 /bin/[*
-r-xr-xr-x  2 root  wheel  43120 Dec 11  2011 /bin/test*

If you run [ 2 -gt 1 ]; echo $? it will print 0 in your terminal, 0 means success everything else means failure. If you call test instead: test 2 -gt 1; echo $? it also print 0.

so if just runs a program and checks the exit code, you can also do:

if true; then
    echo success
fi

if false; then
    echo fail
fi

true; echo $?; false; echo $?

will print: success, 0, 1