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如何在GraphQL中执行突变?

graphql node.js

0

Arkadiusz KaÅkus · 技术社区 · 7 年前

在GraphQL中,我们基本上有两种类型的操作:查询和突变。虽然文档中对查询进行了很好的描述,并且有很多这样的例子,但是我很难理解如何执行变异。突变显然是更新方法。

var express = require("express");
var graphqlHTTP = require("express-graphql");
var graphql = require("graphql");
var inMemoryDatabase = require("./inMemoryDatabase").inMemoryDatabase;
var _ = require("lodash-node");

var userType = new graphql.GraphQLObjectType({
  name: "User",
  fields: {
    id: { type: graphql.GraphQLString },
    name: { type: graphql.GraphQLString }
  }
});

var queryType = new graphql.GraphQLObjectType({
  name: "Query",
  fields: {
    user: {
      type: userType,
      args: {
        id: { type: graphql.GraphQLString }
      },
      resolve: function(parent, { id }) {
        return _.find(inMemoryDatabase, { id: id });
      }
    }
  }
});

var mutationType = new graphql.GraphQLObjectType({
  name: "Mutation",
  fields: {
    user: {
      type: userType,
      args: {
        id: { type: graphql.GraphQLString },
        name: { type: graphql.GraphQLString }
      },
      resolve: function(parent, { id, name }) {
        var index = _.findIndex(inMemoryDatabase, { id: id });
        inMemoryDatabase.splice(index, 1, { id: id, name: name });
        return _.find(inMemoryDatabase, { id: id });
      }
    }
  }
});

var schema = new graphql.GraphQLSchema({
  query: queryType,
  mutation: mutationType
});

var app = express();
app.use(
  "/graphql",
  graphqlHTTP({
    schema: schema,
    graphiql: true
  })
);

var port = 9000;
if (process.env.PORT) {
  port = process.env.PORT;
}

app.listen(port);
console.log("Running a GraphQL API server at localhost:" + port + "/graphql");

内存中的数据库只是一个用户对象数组 {id, name}

var inMemoryDatabase = [
  {
    id: "31ce0260-2c23-4be5-ab78-4a5d1603cbc8",
    name: "Mark"
  },
  {
    id: "2fb6fd09-2697-43e2-9404-68c2f1ffbf1b",
    name: "Bill"
  }
];

module.exports = {
  inMemoryDatabase
};

执行查询以按id获取用户,如下所示:

{
 user(id: "31ce0260-2c23-4be5-ab78-4a5d1603cbc8"){
  name
 }
}

改变用户名会是什么样子?

2 回复 | 直到 7 年前

1

2

Austio 7 年前

嘿,我可能完全不知道你在说什么,但我看待突变的方式是这样的

之后,我从resolve函数返回一些东西,它将满足您在 type

var mutationType = new graphql.GraphQLObjectType({
  name: "Mutation",
  fields: {
    user: {
      // You must return something from your resolve function 
      // that will fulfill userType requirements
      type: userType,
      
      // with these arguments, find the user and update them
      args: {
        id: { type: graphql.GraphQLString },
        name: { type: graphql.GraphQLString }
      },
      // this does the lookup and change of the data
      // the last step of your result is to return something
      // that will fulfill the userType interface
      resolve: function(parent, { id, name }) {
        // Find the user, Update it
        // return something that will respond to id and name, probably a user object
      }
    }
  }
});

然后将其作为上下文,传递一些参数并请求用户返回

mutation updateUser {
  user(id: "1", name: "NewName") {
    id
    name
  }
}

在正常的生产模式中,您通常也会有 errors 它可以返回,以传递失败/未找到的更新的不同状态

2

0

Arkadiusz KaÅkus 7 年前

mutation updateUser {
  user(id: "31ce0260-2c23-4be5-ab78-4a5d1603cbc8", name: "Markus") {
    id
    name
  }
}