InferSchemaType with instance methods

Prerequisites

• I have written a descriptive issue title
• I have searched existing issues to ensure the issue has not already been raised

Issue

So I wanna use the newly added InferSchemaType to remove boilerplate interfaces. However im not sure how would one add InstanceMethods, Virtuals, etc. to the Schema without defining the now inferred DocType. When leaving EnforceDocType and Model as defaults, the inferred document type is { [x: string]: any] }. Unfortunatly, typescript does not allow to skip generic parameters for now (https://github.com/microsoft/TypeScript/issues/10571)

// Schema
const schema = new Schema<
  any,
  mongoose.Model<any, any, any, any, any>,
  {
    test: () => void;
  }
>({
  name: { type: String, required: true },
  email: { type: String, required: true },
  avatar: String,
});

type User = InferSchemaType<typeof schema>;
// InferSchemaType will determine the type as follows:
// type User = {
//   [x: string]: any;
// };

[heilmela]

@iammola i updated the comment, it was a false leftover from a previous edit where i used find(). You are absolutely correct the union with null did produce the error. This works:

UserModel.findOne()
  .exec()
  .then((x) => {
    //  (parameter) x: (mongoose.Document<unknown, any, {
    //     name: string;
    //     email: string;
    //     avatar?: string | undefined;
    // }> & {
    //     name: string;
    //     email: string;
    //     avatar?: string | undefined;
    // } & {
    //     _id: mongoose.Types.ObjectId;
    // } & {
    //     ...;
    // }) | null
    if (x !== null) x.test('a');
  });

[PeterPanen]

Wait i just realized that if i remove the InferSchemaType line in my example, and just let it infer automatically without me doing anything, it includes the instance methods 🥳