Question

我有当前运行良好的index.js文件...但是为了在我的测试策略中使用，我想将其分为2个文件：server.js和app.js

我收到一条错误消息，指出我的app.js不是一个函数。我的编码有什么问题？

index.js

import express from 'express';
import express_graphql from 'express-graphql';
import { buildSchema } from 'graphql';

// Construct a schema, using GraphQL schema language
const schema = buildSchema(`
  type Query {
    message: String
  }
`);

// Root resolver
const root = {
  message: () => 'Hello World!'
};

// Create an express server and a GraphQL endpoint
const app = express();
app.use('/graphql', express_graphql({
  schema: schema,
  rootValue: root,
  graphiql: true
}));

/* eslint-disable no-console */
app.listen(4000, () => console.log('Express GraphQL Server Now running On localhost:4000/graphql'));

分解为：

server.js

import app from './app';

/* eslint-disable no-console */
app.listen(4000, () => {
  console.log('Express GraphQL Server Now running On localhost:4000/graphql');
});

app.js

import express_graphql from 'express-graphql';
import { buildSchema } from 'graphql';
import express from 'express';

 export default function () {

  // Construct a schema, using GraphQL schema language
  const schema = buildSchema(`
    type Query {
      message: String
    }
  `);

  // Root resolver
  const root = {
    message: () => 'Hello World!'
  };

  // Create an express server and a GraphQL endpoint
  const app = express();
  app.use('/graphql', express_graphql({
    schema: schema,
    rootValue: root,
    graphiql: true
  }));

}

console.log

yarn start
yarn run v1.9.4
$ babel-node src/server.js
/Users/yves/Developments/WIP/NODE/nodeGraphQL/src/server.js:10
_app2.default.listen(4000, () => {
              ^

TypeError: _app2.default.listen is not a function
    at Object.<anonymous> (/Users/yves/Developments/WIP/NODE/nodeGraphQL/src/server.js:4:5)
    at Module._compile (module.js:653:30)
    at loader (/Users/yves/Developments/WIP/NODE/nodeGraphQL/node_modules/babel-register/lib/node.js:144:5)
    at Object.require.extensions.(anonymous function) [as .js] (/Users/yves/Developments/WIP/NODE/nodeGraphQL/node_modules/babel-register/lib/node.js:154:7)
    at Module.load (module.js:566:32)
    at tryModuleLoad (module.js:506:12)
    at Function.Module._load (module.js:498:3)
    at Function.Module.runMain (module.js:694:10)
    at Object.<anonymous> (/Users/yves/Developments/WIP/NODE/nodeGraphQL/node_modules/babel-cli/lib/_babel-node.js:154:22)
    at Module._compile (module.js:653:30)
error Command failed with exit code 1.

感谢您的反馈

Answer 1

您的app.js正在导出函数，server.js会导入该函数。但是，您实际上并没有调用该函数。在这种结构下，您的server.js应该看起来像这样：

import app from './app';

app().listen(4000, () => {
    // ....
});

也就是说，您还遇到一个问题，就是正在导出的函数app.js实际上没有返回值，它创建了快速服务器，但没有返回它。因此，您还需要调整app.js，方法是在方法主体的末尾添加一个返回值：

  const app = express();
  // ....

  return app;
}

给个机会吧！

Answer 2

您应该return app在从app.js文件导出的函数中，将其导入到server.js文件中，然后以这种方式app().listen调用该函数。您要返回的应用程序是初始化的const app = express()函数，其中包含服务器要运行的listen()属性。

Answer 3

根据@Elliot和@Ekpin答案的有效代码...（@Elliot在该行的第一个）

app.js

import express_graphql from 'express-graphql';
import { buildSchema } from 'graphql';
import express from 'express';

 export default function () {

  // Construct a schema, using GraphQL schema language
  const schema = buildSchema(`
    type Query {
      message: String
    }
  `);

  // Root resolver
  const root = {
    message: () => 'Hello World!'
  };

  // Create an express server and a GraphQL endpoint
  const app = express();
  app.use('/graphql', express_graphql({
    schema: schema,
    rootValue: root,
    graphiql: true
  }));

  return app;

}

server.js

    import app from './app';

    /* eslint-disable no-console */
    app().listen(4000, () => {
      console.log('Express GraphQL Server Now running On localhost:4000/graphql');
    });
)

如何正确地将index.js拆分为server.js和app.js？

3 个答案: