mjhjmtu

I'm trying to use getops in a function yet it doesn't seem to work:

#!/bin/bash

function main()

 while getopts ":p:t:c:b:" o; do
 case "$o" in
 p)
 echo "GOt P"
 p=$OPTARG
 ;;
 t)
 echo "GOt T"
 t=$OPTARG
 ;;
 c)
 echo "GOt C"
 c=$OPTARG
 ;;
 b)
 echo "GOt b"
 b=$OPTARG
 ;;
 *)
 #usage 
 echo "Unknown Option"
 return 
 ;;
 esac
 done

 echo $p
 echo $t
 echo $c
 echo $b


main

And then running it like this:

$ ./bin/testArguments.sh -p . -t README.md -c 234 -b 1

I have tried making sure that optid are local yet this didn't work either. Anything else that might be wrong?

You're not passing any argument to your main function. If you want that function to get the same arguments as passed to the script, pass them along with:

main "$@"

Instead of:

main

Also relevant to your script:

• difference between “function foo() ” and “foo() ”


• Why is printf better than echo?


• Security implications of forgetting to quote a variable in bash/POSIX shells


• you'd want to output errors on stderr: echo >&2 Unknown option
  


• you'd want to return a non-zero exit status upon error (return 1)


• when calling getopts in a function, it's a good habit to set OPTIND to 1 initially, in case getopts has been called before (for instance in a previous invocation of the function).

In a function, the parameters are those passed to the function, not those passed to the script:

$ cat foo.sh
function main ()

 echo "$@"


echo "$@"
main
$ bash foo.sh bar
bar

$

You need to pass "$@" to main:

main "$@"

Though I find a main function in scripts rather useless, unless you plain to call main again and again in the same script.