Fix: prefer-destructuring says "Use destructuring" on "super" (#9625)

[gjbkz]

What is the purpose of this pull request? (put an "X" next to item)

[ ] Documentation update
[x] Bug fix (template)
[ ] New rule (template)
[ ] Changes an existing rule (template)
[ ] Add autofixing to a rule
[ ] Add a CLI option
[ ] Add something to the core
[ ] Other, please explain:

This fixes #9625

What changes did you make? (Give an overview)

• tests/lib/rules/prefer-destructuring.js: I added a test to reproduce prefer-destructuring says "Use destructuring" on "super" #9625
• lib/rules/prefer-destructuring.js: I updated a line to fix prefer-destructuring says "Use destructuring" on "super" #9625

Is there anything you'd like reviewers to focus on?
No

[not-an-aardvark]

Looks good to me. Thanks for contributing!

[platinumazure]

Could we add a test case for a nested super call?

class Foo extends Bar { static get foo() {var bar = super.foo.bar } }

Asking because I'm not 100% sure that line 148 of the rule will catch that, since you'd have to check rightNode.object.object in that case. In fact, you might need a recursive check, if I'm not mistaken about this. Even if does work, it'd be good to have a test so we can protect against a regression in that case.

[not-an-aardvark]

I agree that it would be nice to have a test. Having said that, I think the implementation in this PR already handles that case correctly by reporting an error because const {bar} = super.foo is legal.

Conceptually, the rule should report an error only when node.object is a valid expression on its own. So I think checking the type of node.object is correct here, regardless of whether node.object is another MemberExpression or contains super.

[platinumazure]

Thanks for the explanation. You are right that I had misunderstood conceptually how this rule is supposed to work and thought that const { bar } = super.foo; was invalid syntax, when in reality it is valid. Thanks!

[platinumazure]

LGTM, thanks!

[gjbkz]

Thank you for the reviewing. I added another test to check what @platinumazure pointed out.
I found that I can omit get and the simplest test reproducing #9625 is

class Foo extends Bar {
  foo() {
    var foo = super.foo;
  }
}

[not-an-aardvark]

Thanks for contributing!