javascript - GraphQL 不允许查询不同的字段

Question

我是 GraphQL 的绝对初学者。如果问题是业余的，请道歉。我正在编写这个 graphQL 端点，它将按姓名过滤人员。效果很好，但我希望能够

按姓名、年龄、电话号码等多个参数过滤

或者

使用参数的组合来搜索..

我目前的程序：

var express = require("express");
var express_graphql = require("express-graphql");
var { buildSchema } = require("graphql");
var Sequelize = require("sequelize")
var db = new Sequelize(
  "users",
  "root",
  "pass",
  {
    host: "localhost",
    dialect: "mysql"
  }
);

var calls = db.define(
  "calls",
  {
    person_id: {
      type: Sequelize.INTEGER,
      allowNull: false,
      primaryKey: true
    },
    name: {
      type: Sequelize.STRING,
      allowNull: false
    },
    age: {
      type: Sequelize.INTEGER,
      allowNull: false
    },
    address: {
      type: Sequelize.STRING,
      allowNull: false
    },
    company: {
      type: Sequelize.String,
      allowNull: false
    },
    email: {
      type: Sequelize.STRING,
      allowNull: false
    }
  },
  {
    tableName: "person"
  }
);

// Construct a schema, using GraphQL schema language
var schema = buildSchema(`
  type Query {
    Person(person_id: Integer!): [Person]
  }
  type Person {
    person_id: Integer
    name: String
    age: Integer
    address: String
    company: String
    email: String 
  }
`);

var getperson = function(args) {
  if (args.person_id) {
    var p_name = args.person_id;
    return calls.findAll({
      where: {
        name: p_name
      }
    });
  }
};

// The root provides a resolver function for each API endpoint
var root = {
  person: getperson
};

var app = express();
app.use(
  "/graphql",
  express_graphql({
    schema: schema,
    rootValue: root,
    graphiql: true
  })
);
app.listen(4000);
console.log("Running a GraphQL API server at http://localhost:4000/graphql");

尝试这样做：

 type Query {
    Person(person_id: Integer!): [Person]
    Person(name: String!): [Person]
  }

但它说 Person 只能定义一次。

标签： javascriptmysqlnode.jsgraphqlsequelize.js

Answer 1

您不能“重载”字段定义。您可以使用可为空的参数定义单个字段

person(personId: Int, name: String)

或创建不同名称的字段

personById(id: Int!)
personByName(name: String!)