composer-json argument needs unused command to be passed

[icanhazstring]

Describe the bug

Trying to use composer-unused on a global binary and passing along a certain composer.json has a really awkward usage.

You have to use:

composer-unused.phar unused <path-to-composer.json>

The unused command name should not be present at all.
This is due to the "multi command" console application.

https://github.com/composer-unused/composer-unused/blob/main/bin/composer-unused#L60-L64
Maybe there is a better way to do that.

[MarcinGladkowski]

Hi!

PR: #608

In my opinion this is happen because of limitation of implementation default command in symfony/console
https://github.com/symfony/console/blob/f8587c4cdc5acad67af71c37db34ef03af91e59c/Application.php#L1193-L1204

My idea to resolve it is:

• while any of rest of commands aren't used as first argument, I mean
  debug:consumed-symbols or 'debug:provided-symbols' the package treats unused command as default single command.

$application->setDefaultCommand('unused', true);

That change behaviour from

./bin/composer-unused composer.json         

Command "composer.json" is not defined.                                            

To:

Results
-------

Found 17 used, 0 unused, 0 ignored and 1 zombie packages

But while using other commands is still works as expected:

./bin/composer-unused debug:consumed-symbols                                                                                       1 err | system node | 
ArrayIterator
Closure
// ...

Testing:
I'm testing it with generating phar locally, but write PHPunit integration tests is quite not conventional. I cannot easily get output even using buffer.