Sequelize 连接表而不选择

Question

我想从 Sequelize 中多对多关系的连接表中选择 ID，并根据端点之一限制结果。例如：

ModelA:
- id
- deleted_at
- is_expired

ModelB:
- id
- deleted_at
- is_active

ModelAToModelB:
- a_id
- b_id

我想做类似的事情

SELECT id FROM ModelAToModelB ab INNER JOIN ModelB b ON ab.b_id = b.id WHERE ab.id = someA_ID AND b.deleted_at IS NULL;

我目前正在做类似的事情

const ModelB = require("./modelb")
const ModelAToModelB = require("./modelatob");

ModelAToModelB.findAll({
    where: { id: someA_ID },
    include: [
       {
           model: ModelB,
           where: { deleted_at: null }
       }
    ]
})

这似乎有效，但除此之外ModelB，当我想要的只是ModelAToB.id.

有什么办法可以报废ModelB数据吗？或者可能使用 fromModelA来获取关联 ID？

Answer 1

const ModelB = require("./modelb")
const ModelAToModelB = require("./modelatob");

ModelAToModelB.findAll({
    where: { id: someA_ID },
    include: [
       {
           model: ModelB.scope(null), //Prevent the default scope from adding attributes
           where: { deleted_at: null },
           required: true, //Force an inner join
           attributes: [] //Join without selecting any attributes
       }
    ]
})