mysql - 如何使用以下和跟随者示例在 sequelize 中加入两次单个表

Question

这段代码运行完美。我现在可以获得关注和关注者，我需要查看我获得的关注者是否也在关注他们？

问题是我如何对 Followers 表进行另一个查询/子查询并查看我是否也在关注我的关注者。

追随者表

export default function(sequelize, DataTypes) {
  return sequelize.define('Follower', {
    _id: {
      type: DataTypes.INTEGER,
      allowNull: false,
      primaryKey: true,
      autoIncrement: true
    },
    userId: {
      type: DataTypes.INTEGER,
      allowNull: false
    },
    followingId: {
      type: DataTypes.INTEGER,
      allowNull: false
    }
  });
}

协会

db.Follower.belongsTo(db.User, {as:'following', foreignKey: 'followingId'});
db.Follower.belongsTo(db.User, {as:'follower', foreignKey: 'userId'});

查询

Follower.findAll({
        where: {
            followingId: userId
        },
        attributes: ['_id'],
        include: [
          {
            model: User,
            attributes: ['fullName', 'username', '_id', 'picture'],
            as: 'follower'
          }
        ]
    })

更新

我已经从行查询中获得了期望的结果:

SELECT  F.userId, F.`followingId` , F1.`followingId` as IsFollowing , U.`fullName` FROM Followers as F

INNER JOIN Users as U ON userId = U._id

LEFT JOIN Followers as F1 On F.userId = F1.followingId

WHERE F.followingId = 142 

仍在 Sequelize 中挣扎。

Answer 1

将您的行查询转换为 sequlize 请求试试这个:

协会

db.Follower.belongsTo(db.User, {as: 'following', foreignKey: 'followingId', sourceKey: 'userId'});
db.Follower.hasOne(db.User, {as: 'follower', foreignKey: 'userId'});

查询

Follower.findAll({
  where: {
    followingId: userId
  },
  attributes: ['userId', 'followingId'],
  include: [
    {
      model: User,
      attributes: ['fullName'],
      as: 'follower',
      required: true // to get inner join
    },
    {
      model: Follower,
      attributes: [['followingId', 'IsFollowing']],
      as: 'following',
      required: false // to get left join
    }
  ]
});