Mongoose + Typescript -> Exporting model interface
Try:
interface UserDocument extends IUser, Mongoose.Document { _id: string;}
It will resolve the conflict between IUser._id (string) vs Mongoose.Document._id (any).
Update:
As pointed out in comments, currently it gives a incompatible override for member from "Document"
, so another workaround must be used. Intersection types is a solution that can be used. That said, the following can be done:
type UserDocument = IUser & Mongoose.Document;
Alternatively, if you do not want UserDocument
anymore:
// Modellet Users = Mongoose.model<IUser & Mongoose.Document>('User', userSchema);
It is worth noting that there is a side effect in this solution. The conflicting properties will have the types intersected, so IUser._id (string) & Mongoose.Document._id (any)
results in UserDocument._id (any)
, for example.
I just had this exact issue, where I wanted to keep the User interface properties as separate from Mongoose as possible. I managed to solve the problem using the Omit utility type.
Here is your original code using that type:
import { Document, Model, ObjectId } from 'mongoose';export interface IUser { _id: ObjectId; name: string; email: string; created_at: number; updated_at: number; last_login: number;}export interface IUserDocument extends Omit<IUser, '_id'>, Document {}export interface IUserModel extends Model<IUserDocument> {}
try this:
const UserSchema: Schema = new Schema( { .. });type UserDoc = IUser & Document;export interface UserDocument extends UserDoc {}// For modelexport interface UserModel extends Model<UserDocument> {}export default model<UserDocument>("User", UserSchema);