Sequelize error when using “where” and “in” on a subarray

Question

This is my model definition:

var Tag = sequelize.define(\'Tag\', {
    name: Sequelize.STRING
});

var Event = sequelize.define(\'Event\', {
    name: Sequelize.         


        

Answer 1

replace in with $in

where : { id : { $in : [1,2,3,4]}}

Answer 2

in property able to use at version 4.42

Tag.findAll({
    where: { id: {in: [1,2,3,4]} }
}).then(...)

And you can use notIn property for excluded values.

Tag.findAll({
    where: { id: {notIn: [1,2,3,4]} }
}).then(...)

http://docs.sequelizejs.com/manual/querying.html#operators

Answer 3

I know this is hella old. But I'm pretty sure what you were looking for is as follows. At least one of these should do the trick. Probably both. I can't test at the moment because I'm in the middle of a 43-files-changed spaghetti code beast. I'll update with an actual test when I find the time.

(I'm posting now because all the rest seem to not actually answer the question asked, but they did answer the question I had when I arrived here.)

Event
.findAndCountAll({
    include: [{model: Tag, as: 'tags'}],
    where: {
      tags: {
        id: {in: [1,2,3,4]}
      }
    },
    order: order,
    limit: pageSize,
    offset: pageSize * (page - 1),
})
.success(function(result) {

    ...
});

Event
.findAndCountAll({
    include: [
      {
        model: Tag,
        as: 'tags',
        where: { id: {in: [1,2,3,4]} },
      }
    ],
    order: order,
    limit: pageSize,
    offset: pageSize * (page - 1),
})
.success(function(result) {

    ...
});

Answer 4

No need "in" property, sequelize auto define this. Just set array.

Tag.findAll({
    where: {
        id: [1,2,3,4]
    }
}).then(...)

Answer 5

Updated example for Sequelize v5:

await Tag.findAll({
  where: {
    id: {
      [Sequelize.Op.in]: [1, 2, 3, 4]
    }
  }
});