node.js - 如何将结果列表从 sequelizejs 返回到 grpc？

Question

如何使用 grpc 返回用户列表，结果为空:

我是http://condorjs.com , 创建 grpc 服务器

class User {
  getUser(ctx) {
    models.users.findAll({
            where: { userId: ctx.req.userId }
        }).then(function(users){
            // return a list of users
           for (var i = 0, len = users.length; i < len; i++) {
             result.push({'userId': users[i].userId, 'userName': users[i].userName);
           }

           return { 'users': result};
        });
     }
   }

const options = {
      'listen': '0.0.0.0:50051',
      'rootProtoPath': 'protos',
   };

   const app = new Condor(options)
    .add('user.proto', 'User', new User())
    .start();

GRPC 客户端调用:

var PROTO_PATH = __dirname + '/../proto/profile.proto';

var grpc = require('grpc');
var user_proto = grpc.load(PROTO_PATH).user;

function main() {
  var client = new guser_proto.User('localhost:50051',
                                       grpc.credentials.createInsecure());

    client.getUser({userId: '8888'}, function(err, response) {
      console.log(response);
    });
}

main();

原型(prototype):

message UserResponse {
    message User{
        string userId = 1;
        string userName = 2;
    }

    repeated User users= 1;
}

Answer 1

你在函数定义中没有收到任何回调方法，所以我添加了一个回调参数，这个回调方法将使用所需的参数调用。您将在 main 方法中收到数据

 getUser(ctx, callback) {
    models.users.findAll({
            where: { userId: ctx.req.userId }
        }).then(function(users){
            // return a list of users
           for (var i = 0, len = users.length; i < len; i++) {
             result.push({'userId': users[i].userId, 'userName': users[i].userName);
           }

           callback(null,{ 'users': result});
        }.bind(this));
     }
   }

这肯定会在这里返回您的数据

function main() {
  var client = new guser_proto.User('localhost:50051',
                                       grpc.credentials.createInsecure());

    client.getUser({userId: '8888'}, function(err, response) {
      console.log(response); // it will print here 
    });
}