如何从需要返回语句的GraphQL解析器中调用异步node.js函数？创建一个简单的GraphQL实现提供<a href="http://graphql.org/graphql-js" rel="nofollow noreferrer">graphql.org/graphql-js</a>

Question

的Hello World示例如下：

var { graphql, buildSchema } = require('graphql'); 

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

// The root provides a resolver function for each API endpoint 
var root = { 
    hello:() => { 
     return 'Hello World!'; 
    } 
}; 

// Run the GraphQL query '{ hello }' and print out the response 
graphql(schema, '{ hello }', root).then((response) => { 
    console.log(response); 
}); 

我试图运行变压器内的异步函数，例如数据库调用，我不能似乎弄清楚如何使这项工作：

我想要做的事：

var { graphql, buildSchema } = require('graphql'); 

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

// The root provides a resolver function for each API endpoint 
var root = { 
    data:() => { 
     getData((data) => { 
      return data; // Returns from callback, instead of from resolver 
     } 
    } 
}; 

// Run the GraphQL query '{ data }' and print out the response 
graphql(schema, '{ data }', root).then((response) => { 
    console.log(response); 
}); 

我试图用承诺，但似乎没有要逃避的一种方式承诺没有ent进行回调。我也尝试过用各种方法来封装异步函数，强制它是同步的，但最终不得不从异步函数中返回一个值，同样的问题。我觉得这不能太复杂，我的意思是GraphQL-JS是由Facebook编写的。

Answer 1

所以这个问题原来是你在解决问题后觉得非常愚蠢的问题之一，但是因为我花了多长时间来解决问题，我会回答我自己的问题，以便其他人能够希望得到好处。

事实证明，GraphQL解析器函数必须返回一个值或者解析为该值的承诺。所以我试图用像这样的事：

var { graphql, buildSchema } = require('graphql'); 

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

// The root provides a resolver function for each API endpoint 
var root = { 
    data:() => { 
     getData((data) => { 
      return data; // Returns from callback, instead of from resolver 
     } 
    } 
}; 

// Run the GraphQL query '{ data }' and print out the response 
graphql(schema, '{ data }', root).then((response) => { 
    console.log(response); 
}); 

可与东西就像这样：

var { graphql, buildSchema } = require('graphql'); 

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

let promiseData =() => { 
    return new Promise((resolve, reject) => { 
     getData((data) => { 
      resolve(data); 
     }); 
    }); 
}; 

// The root provides a resolver function for each API endpoint 
var root = { 
    data:() => { 
     return promiseData(); 
    } 
}; 

// Run the GraphQL query '{ data }' and print out the response 
graphql(schema, '{ data }', root).then((response) => { 
    console.log(response); 
});