Typescript: Model.findAll returns any

import { Sequelize, Model, DataTypes } from 'sequelize'

const sequelize = new Sequelize('postgres://user:password@localhost:5432/db')

export class User extends Model {
  public id?: number
  public name!: string
}

User.init(
  {
    name: {
      type: DataTypes.STRING,
      allowNull: false,
    },
  },
  {
    sequelize,
    tableName: 'user',
  }
)

async function test() {
  const users = await User.findAll()
}

users is typed as any. I would have expected it to be of type User[]. I use sequelize 5.8.5 and typescript 3.4.5.

It seems that the static definition of findAll is overrided by a method.

[upcyoung]

see typescript install

@types/node (this is universally required) @types/validator @types/bluebird

After npm install @types/validator @types/bluebird, my question is resolved.

[Andrew1431]

Oops sorry I thought your response was for another issue.

If you’re on <v6 of sequelize, you need to add bluebird types as a dependency. Otherwise upgrade sequelize to v6. That is what their current documentation represents regarding only needing validators and node types