Question

我正在使用graphql，sequelize和graphql-sequelize，而且我遇到了一些麻烦来解决"IS A"关系。

我的续集模型如下：

// models.js

// User table
let User = sequelize.define('user', {
    id: Sequelize.INTEGER
    name: Sequelize.STRING
});

// Patient table
let Patient = sequelize.define('patient', {
    bloodType: Sequelize.STRING
});
// Defines "IS A" relationship: [Patient] IS A [User]
Patient.belongsTo(models.User, {
    foreignKey: {
      name: 'id',
      type: DataTypes.INTEGER,
      primaryKey: true
    },
    foreignKeyConstraint: true
})

// Doctor table
let Doctor = sequelize.define('doctor', {
  registry: Sequelize.STRING
});
// Defines "IS A" relationship: [Doctor] IS A [User]
Doctor.belongsTo(models.User, {
    foreignKey: {
      name: 'id',
      type: DataTypes.INTEGER,
      primaryKey: true
    },
    foreignKeyConstraint: true
})

这是我的graphql架构：

// graphql.js
import resolver from 'graphql-sequelize';
import * as models from './models';

let userType = new GraphQLObjectType({
    name: 'User',
    fields: {
        id: { type: new GraphQLNonNull(GraphQLInt) },
        name: { type: GraphQLString }
    }
});

let patientType = new GraphQLObjectType({
    name: 'Patient',
    fields: {
        id: { type: new GraphQLNonNull(GraphQLInt) },
        bloodType: { type: GraphQLString },
        user: {
            type: new GraphQLNonNull(userType),
            // IMPORTANT!
            // How can I call resolver if I don't have a assotiation property like Patient.User?
            resolve: resolver()
        }
    }
});

// [doctorType omitted]

let schema = new GraphQLSchema({
  query: new GraphQLObjectType({
    name: 'Query',
    fields: {
        users: {
            type: userType,
            resolve: resolver(models.User)
        },
        patients: {
            type: patientType,
            resolve: resolver(models.Patient)
        }
        // [doctor field omitted]
    }
  })
});

在架构（resolver和users）字段上调用patients方法可以正常工作，但我的问题是如何在patientType中调用解析器返回它的用户，因为我没有关联财产。

感谢。

Answer 1

正如mickhansen在issue上提到的，所有续集关联调用都会返回一个关联引用。

我存储了从belongsTo返回的引用，如上所示：

 models.Patient.User = Patient.belongsTo(models.User, {
    foreignKey: {
      name: 'id',
      type: DataTypes.INTEGER,
      primaryKey: true
    },
    foreignKeyConstraint: true
})

然后在resolve函数中使用它：

import * as models from '../path/models'

let patientType = new GraphQLObjectType({
    name: 'Patient',
    fields: {
        id: { type: new GraphQLNonNull(GraphQLInt) },
        bloodType: { type: GraphQLString },
        user: {
            type: new GraphQLNonNull(userType),
            resolve: resolver(models.default.Patient.User)
            // alternatively you could the following as sequelize always stores assotiations in associations
            // resolve: resolver(models.default.Patient.associations.User)
        }
    }
});

解决＆＃34; IS A＆＃34;与graphql-sequelize的关系

1 个答案: