检查AST是否为TSLint自定义规则的@Injectable装饰器

Question

I am trying to design a custom TSLint rule that will allow for me to check and make sure that the @Injectable decorator is not included within classes that do not inject any other services into themselves via the constructor.

I have been reading this documentation and have been able to write a rule previously to disallow export modifiers in test files by doing:

const hasExport = node.modifiers && node.modifiers.some(
    (modifier) => modifier.kind === SyntaxKind.ExportKeyword
);

This AST explorer website has been helpful throughout this process but I am having trouble deciding the best way to figure out if the @Injectable decorator is being added to my node.

So far I have:

const hasInjectableDecorator = node.decorators && node.decorators.some(
    (decorator) => decorator.kind === SyntaxKind.Decorator
);

But this will only check to see if it is of any decorator (@Injectable, @Component, @NgModule, etc.) and I was wondering how I could then check the text, using the available properties/methods on decorator to make sure it is indeed an @Injectable . I would like to start with this first and then dive into ensuring no other services are being injected.

Thanks

Answer 1

+1 for astexplorer.net! Great website.

import * as ts from "typescript";

const hasInjectableDecorator = (node: ts.ClassDeclaration) => node.decorators
    && node.decorators.some(isInjectableDecorator);

const isInjectableDecorator = (decorator: ts.Decorator) => ts.isIdentifier(decorator.expression)
    && decorator.expression.text === "Injectable"

Explanation: you want to check whether the decorator's expression has a text value of "Injectable" , right? This method will check if the decorator is an Identifier (it refers to a variable), and if so, checks what text name it uses.

Fun fact: instead of checking node.kind === ts.SyntaxKind.Decorator , you should use the built-in ts.isDecorator method (and its equivalents for other node syntax kinds) . They directly check .kind the same way and also return node is ts.Decorator , so you'll get type inference without having to manually cast node as ts.Decorator later on.