Question

我有三个组件A，B和C。当A向B发送HTTP请求时，B会向C发送另一个HTTP请求，检索相关内容并将其发送回A，作为HTTP响应。

B组件由以下Node.js代码段表示。

var requestify = require('requestify');

// sends an HTTP request to C and retrieves the response content
function remoterequest(url, data) {
  var returnedvalue;
  requestify.post(url, data).then(function(response) {
    var body = response.getBody();
    // TODO use body to send back to the client the number of expected outputs, by setting the returnedvalue variable
  });
  return returnedvalue;
}

// manages A's request
function manageanotherhttprequest() {
  // [...]
  var res = remoterequest(url, data);
  // possible elaboration of res
  return res;
}

由于body功能，我需要返回remoterequest个内容。我注意到，目前，POST请求是异步的。因此，在将returnedvalue变量返回给调用方法之前，永远不会分配它。

如何执行同步HTTP请求？

Answer 1

您正在使用restify，一旦调用其方法（promise，post等等），它将返回get。但是，您创建的remoterequest方法未返回promise，您无法使用.then等待。您可以使用promise或内置async-await返回promise，如下所示：

使用承诺：

var requestify = require('requestify');

// sends an HTTP request to C and retrieves the response content
function remoterequest(url, data) {
  var returnedvalue;
  return new Promise((resolve) => {
    requestify.post(url, data).then(function (response) {
      var body = response.getBody();
      // TODO use body to send back to the client the number of expected outputs, by setting the returnedvalue variable
    });
    // return at the end of processing
    resolve(returnedvalue);
  }).catch(err => {
    console.log(err);
  });
}

// manages A's request
function manageanotherhttprequest() {
  remoterequest(url, data).then(res => {
    return res;
  });
}

使用async-await

var requestify = require('requestify');

// sends an HTTP request to C and retrieves the response content
async function remoterequest(url, data) {

  try {
    var returnedvalue;
    var response = await requestify.post(url, data);
    var body = response.getBody();
    // TODO use body to send back to the client the number of expected outputs, by setting the returnedvalue variable
    // return at the end of processing
    return returnedvalue;
  } catch (err) {
    console.log(err);
  };
}

// manages A's request
async function manageanotherhttprequest() {
    var res = await remoterequest(url, data);
    return res;
}

Node.js

1 个答案: