Variable Syntax Error with Bash Script (Homework)

Question

I\'m currently working on homework for a Unix course and I\'ve got everything else done but I just can\'t figure out what I\'ve done wrong here. I\'ve done this exact same s

Answer 1

Update: As the OP discovered, another issue was a malformed shebang line: #/!bin/bash instead of #!/bin/bash (consider using #!/usr/bin/env bash, though).

A tip beforehand: http://shellcheck.net is a very handy tool for checking shell code for errors.

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@Ube and @kojiro's answers both contain important pointers:

• ' is used for quoting, so it must either be used inside double quotes, or it must be escaped as \'.

• No spaces are allowed around the = in variable assignments in bash.

But there's more:

• Use $1 to refer to the first argument; $(1) does something fundamentally different: it uses command substitution ($(...)) to execute the enclosed command and return its stdout output; thus, $(1) tries to execute a - most likely non-existent - command 1 and return its output.

• Always use [[ ... ]] rather than [ ...] in bash; it's more robust and provides more features; case in point: [ $input != 5 ] will break if $input is undefined or empty - you'd have to double-quote it to prevent that; by contrast, it's fine to use [[ $input != 5 ]].

• All case branches must be terminated with ;; (even the branches that just contain exit commands;).^{Fine print: the very last branch also works without ;;}

• The if [ -d $finput ] ; then ... statement is missing its closing fi.