Question

A.js ：

A.associate = models => {
 A.hasMany(models.B, { foreignKey: 'A_id' });
 A.hasMany(models.C, { foreignKey: 'A_id' });
 A.hasMany(models.D, { foreignKey: 'A_id' });
};

A.findOne({
 where: { id: _id },
 include: [ B, C, D ],
});

B.js ：

B.associate = models => {
 B.belongsTo(models.A, { foreignKey: 'A_id' });
};

C.js ：

C.associate = models => {
 C.belongsTo(models.A, { foreignKey: 'A_id' });
};

D.js ：

D.associate = models => {
 D.belongsTo(models.A, { foreignKey: 'A_id' });
};

DB ：

A TABLE
 id:1 name: A
B TABLE
 A_id:1 name: B1
 A_id:1 name: B2
blabla...

A :: findOne的结果：

A
 id:1 blabla...
B
 A_id:1 name: B1
 A_id:1 name: B2 -> NO FIND B2

我要搜索“单个A记录”和“多个B，C，D记录”。我怎样才能做到这一点？续集中的关联是如此混乱。

感谢您的帮助！

@@@@@@来源

routes / A.js

router.get('/:id', (req, res) => {
  const { A } = require('../models');
  A.getById(req.params.id)
    .then(record => {
      console.log(record[0].Bs);
    });
});

models / A.js

A.associate = models => {
    A.belongsTo(models.User, {foreignKey: 'userId'});
    A.hasMany(models.B, {foreignKey: 'A_Id'});
    A.hasMany(models.C, {foreignKey: 'A_Id'});
  };

  A.getById = _id => {
    const { B, C } = require('../models');
    return A.findAll({
      where: { id: _id },
      include: [ B, C ],
    });
  };

models / B.js

B.associate = models => {
    B.belongsTo(models.A, { foreignKey: 'A_Id' });
  };

执行的查询

SELECT `A`.`id`, `A`.`userId`, `A`.`name`,
`Bs`.`A_Id` AS `Bs.A_Id`, `Bs`.`name` AS `Bs.name`,
`Cs`.`A_Id` AS `Cs.A_Id`, `Cs`.`name` AS `Cs.name`
FROM `As` AS `A`
LEFT OUTER JOIN `Bs` AS `Bs` ON `A`.`id` = `Bs`.`A_Id`
LEFT OUTER JOIN `Cs` AS `Cs` ON `A`.`id` = `Cs`.`A_Id`
WHERE `A`.`id` = '1'

登录路线/A.js

[ B {
  dataValues: { A_Id: 1, name: 'Name1' },
  _previousDataValues: ...
  ...
} ]

但是执行查询的结果是多行。（在数据库客户端程序中）

Nodejs Sequelize与关联查找

@@@@@@来源

0 个答案: