CODEWITHSUNDEEP
joinsequelize.jsfieldalias
U4EA
U4EA

Reputation: 912

Sequelize query - is it possible to alias field in included (inner join) model

Can't seem to find an answer to this (if there is one).

So I have 2 sequelize models. The association is working fine but I wish to alias a field from the outer model on which the inner is being performed.

When I performed this query: -

ModelOne.findAll({
    attributes: ['id', ['attribute2', 'attribute2Alias'],],
    include: [{
        model: ModelTwo,
        required: true,
        attributes: [
            ['attribute1', 'attribute1Alias']
        ],
    }],
})

I get back this: -

items: [{
    id: 1,
    attribute2Alias: 'value',
    'ModelTwoTableName.attribute1Alias': 'value'
}]

I understand that I could use .map on the the returned rows but is there any way to avoid this by performing an alias directly?

Thanks.

Upvotes: 0

Views: 5856

Answers (3)

Maksym Dudyk
Maksym Dudyk

Reputation: 1163

In stable 6.0 version of Sequelize the following works out:

ModelOne.findAll({
...
include: {association: 'modeltwos', attributes: []}
})

Note: 'modeltwos' is alias of ModelTwo, set in ModelOne model:

static associate({ ModelTwo }) {
      // define association here
      models.ModelOne.hasMany(ModelTwo, { as: 'modeltwos' }); 
}

Upvotes: 0

Bob Siefkes
Bob Siefkes

Reputation: 1413

I run into the same shortcoming and found the solution. You could refer to the outer column attribute from the top level attributes list. Then use sequelize.col() together with the table name. This requires raw:true otherwise sequelize will complain. The outer attributes list could be left empty. Like this:

ModelOne.findAll({
    attributes: ['id', ['attribute2', 'attribute2Alias'], [sequelize.col('table2.attribute1'), 'attribute1Alias']],
    include: [{
        model: ModelTwo,
        required: true,
        attributes: [],
    }],
    raw: true,
})

Upvotes: 1

Vivek Doshi
Vivek Doshi

Reputation: 58593

It only happens when you use 'raw:true' in your query ,

I think the output you are showing is with raw:true , remove that and then run your query again

Upvotes: 4

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