Question

我有一张自助人士的餐桌，所以人们可以有父母/孩子/堂兄弟等。

const People = sequelize.define('People', {
  gender: Sequelize.STRING,
  name: Sequelize.STRING,
  age: Sequelize.INTEGER
})

const Relationships = sequelize.define('Relationships')
Items.belongsToMany(Items, { through: Relationships, as: 'relationships' })

我希望能够以两种方式选择数据：

1。选择年龄为21岁的人的所有关系

// Returns all of johns relatives who are 21
return People.findOne({
  where: { name: 'John' },
  include: [{
    required: false,
    model: Items,
    as: 'relationships',
    where: { age: 21 }
  }]
})

2。选择所有年龄为21岁的关系的人。这需要接受多个查询，例如：选择所有具有21或/和男性亲属的人。

有什么想法吗？

Answer 1

这里有一些完整的玩具代码，我希望它对某人有用。请注意，在这个例子中，关系不是互惠的，这意味着如果约翰与玛丽有关系，那么玛丽也不会自动地与约翰建立关系（这更像约翰遵循玛丽的情况）。但它仍然是如何与显式连接表进行自我关联的一个例子。

let Sequelize = require('sequelize');
let sequelize = new Sequelize('test', 'test', 'test', {dialect: 'mysql'});

let Person = sequelize.define('Person', {
    name: Sequelize.STRING,
    gender: Sequelize.STRING,
    age: Sequelize.INTEGER
});

let PersonRelationship = sequelize.define('PersonRelationship' /* , more fields could be defined here */);

Person.belongsToMany(Person, {as: 'Relationships', through: PersonRelationship, foreignKey: 'from'});
Person.belongsToMany(Person, {as: 'ReverseRelationships', through: PersonRelationship, foreignKey: 'to'});

let john, mary;

sequelize.sync()
    .then(() => Person.create({name: 'John', gender: 'm', age: 25}))
    .then(p => john = p)
    .then(() => Person.create({name: 'Mary', gender: 'f', age: 21}))
    .then(p => mary = p)
    .then(() => john.addRelationship(mary))
    .then(() => john.getRelationships({where: {age: 21}}))
    .then(relationships => {
        for (let relationship of relationships) {
            console.log('Found relationship:', relationship.name);
        }
    });

基于Sequelize中包含的数据进行查询

1 个答案: