Node.js“ q”包异步进行同步[重复]

Question

这个问题已经在这里有了答案：

• 如何从异步调用返回响应？ 36个答案

我正在使用q包将我的js代码从异步转换为同步。 我不知道我的理解是否正确。 我写了下面的代码片段

 var Q = require('q') function onMessage() { console.log("Before calling Q.fncall"); Q.fcall(processMessage) .then(function () { console.log("processMessage execution compleated"); }) .done(); console.log("Calling Q.fncall Compleated"); return true; } function processMessage() { var defer = Q.defer(); console.log("processMessage Entered"); setTimeout(function () { console.log("processMessage processed"); defer.resolve(); console.log("processMessage defer Resolved"); }, 10) console.log("ProcessMessage Before Exit"); return defer.promise; } onMessage() 

代码的输出为:::: Before calling Q.fncall Calling Q.fncall Compleated processMessage Entered ProcessMessage Before Exit processMessage processed processMessage defer Resolved processMessage execution compleated

但是我的期望是Before calling Q.fncall processMessage Entered ProcessMessage Before Exit processMessage processed processMessage defer Resolved processMessage execution compleated Calling Q.fncall Compleated获得:::: Before calling Q.fncall processMessage Entered ProcessMessage Before Exit processMessage processed processMessage defer Resolved processMessage execution compleated Calling Q.fncall Compleated

有什么我想念的东西或如何实现....

Answer 1

这应该解决

var Q = require('q')

function onMessage() {
  console.log("Before calling Q.fncall");
  return new Promise(resolve => {
    Q.fcall(processMessage)
      .then(function () {
        console.log("processMessage execution compleated");
        console.log("Calling Q.fncall Compleated");
      })
      .done(() => resolve(true));
  })
}

function processMessage() {
  var defer = Q.defer();
  console.log("processMessage Entered");
  setTimeout(function () {
    console.log("processMessage processed");
    defer.resolve();
    console.log("processMessage defer Resolved");
  }, 10)
  console.log("ProcessMessage Before Exit");
  return defer.promise;
}

onMessage();