Op.any does not work when typeValidation = true

[simonbrent]

What are you doing?

const sequelize = new Sequelize('database', 'username', 'password', { typeValidation: true });
const model = sequelize.define('foo', {
  id: { type: Sequelize.INTEGER }
});
model.findAll({ where: { id: { [Sequelize.Op.any]: [1,2] } } }).then(console.log);

What do you expect to happen?

Records 1 and 2 are logged

What is actually happening?

Unhandled rejection SequelizeValidationError: [1,2] is not a valid integer
at INTEGER.validate (node_modules/sequelize/lib/data-types.js:175:11)
at Object.escape (node_modules/sequelize/lib/dialects/abstract/query-generator.js:917:26)
at Object._whereParseSingleValueObject (node_modules/sequelize/lib/dialects/abstract/query-generator.js:2387:49)
...

This is because the escape function in query-generator requires options.isList to be true for an array value to be correctly validated, but on line 2387 it is called without any options: this.escape(value, field).
The same is true for Op.all. On lines 2418 and 2422, escape is called for Op.any/Op.all with isList = true, but these lines only apply to ANY/ALL conditions which are nested, e.g.

{ where: { [Sequelize.Op.not]: { id: { [Sequelize.Op.any]: [ 1, 2 ] } } }

Dialect: postgres
Dialect version: 10
Sequelize version: 4.42.0
Tested with latest release: Yes

Note : Your issue may be ignored OR closed by maintainers if it's not tested against latest version OR does not follow issue template.

[reezer]

This is not just related to typeValidation cases. This is a fixed minimal example (id missing in the above):

const model = sequelize.define('user', {
  id: {
    type: Sequelize.INTEGER,
    primaryKey: true,
  }
}, {
    tableName: 'users',
    underscored: true,
  });

model.findOne({
  where: {
    id: { [Sequelize.Op.any]: [1, 2] }
  }
}).then(console.log);

In 4.37.10 this is still working.

[simonbrent]

Urgh, sorry about that. Fixed my initial comment.