Have options independent of module about vuex-class-modules HOT 3 CLOSED

The simplest implementation would be having VuexModule as:

export class VuexModule {
  private __options: RegisterOptions;
  constructor(options: RegisterOptions | ((className: string) => RegisterOptions)) {
    if (typeof options === "function") {
      options = options(this.constructor.name);
    }
    this.__options = options;
  }
}

The problem with using this.constructor.name is that in the minified version it will be mangled. Maybe multiple different classes might even get the same name then⁉

from vuex-class-modules.

I don't see any use cases other than DI where you wouldn't want to add a name when registering the modules (for instance in Vue Devtools you would like to be able to identify the modules). Vuex' store.registerModule also requires a name (path).

As you mention, using constructor.name has the problem that it will be mangled when minified, and I'm not sure the names will be unique, so I don't like this solution.

For your need, can't you just change the constructor of your modules to something like:

@Module
class MyModule extends VuexModule {
  constructor(store: Store) {
    super({ store, name: "myModule" })
  }
}

?

from vuex-class-modules.

This might not pose an immediate problem, but it goes against the principles of dependency injection, because then every module must make sure not to collide with any other.
When I said, using the module class, I meant “based on the module class”. So some uid could be added.

Anyway, I can of course define my own constructor, which allows the options factory and calls it before invoking super. So it’s ok that this is not changed here.

from vuex-class-modules.