Tips for retrieving a flag when there is a preexisting record within an association in Sequelize

I am working with a model A that has a 1:N association with a model B. My objective is to retrieve all records from A and determine whether there is at least one associated record from B (true) or not (false).

The relationship setup:

ModelA.hasMany(ModelB)
ModelB.belongsTo(ModelA)

The query I currently have is:

const results = await ModelA.findAll({
include: { model: ModelB, required: false }
})

This allows me to retrieve matching records from B when they exist due to the outer join, but my main concern is simply to identify if any records from B are present. Is there a way to customize the SELECT query to directly return true/false based on the result of the join?

My desired custom query would be:

SELECT a.*, CASE WHEN b.id IS NOT NULL THEN true ELSE false END
FROM modelA a LEFT OUTER JOIN modelB b on b.model_a_id = a.id

Answer №1

Sequelize currently does not have built-in support for CASE WHEN, so you will need to utilize the literal function instead.

const results = await ModelA.findAll({
  attributes: {
    include: [
      [Sequelize.literal('CASE WHEN "b"."id" IS NOT NULL THEN true ELSE false END'), 'b_exists']
    ]
  },
  include: { 
    model: ModelB,
    attributes: [],  // add this to remove ModelB data in response
    required: false
  }
})

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