Q承诺的串行执行 [英] Serial execution with Q promises

问题描述

我认为我误解了 Q 承诺如何运作。我希望我的第一个承诺在下一个开始之前解决，但那不会发生。这是我的代码：

I think I'm misunderstanding how Q promises work. I want my first promise to resolve before the next one starts, but that's not happening. Here is my code:

var Q = require('q');

function doWork(taskName) {
  var deferred = Q.defer();
  console.log('starting', taskName);
  setTimeout(function() { 
    console.log('done with', taskName);
    deferred.resolve(); 
  });

  return deferred.promise;
}

doWork('task one')
  .then(doWork('task two'))
  .then(function() { console.log('all done'); });

此代码生成：

$ node test.js 
  starting task one
  starting task two
  done with task one
  done with task two
  all done

我希望它能产生：

$ node test.js 
  starting task one
  done with task one
  starting task two
  done with task two
  all done

我做错了什么？

推荐答案

这有效：

doWork('task one')
  .then(function() {
    return doWork('task two')
  })
  .then(function() {
    console.log('all done'); 
  });

这是有道理的 - 只需直接调用 doWork 在然后（）将立即触发超时，而不是让Q有机会等到任务1 完成。

That makes sense - just calling doWork directly in then() will fire off the timeout immediately, instead of giving Q a chance to wait until task one is complete.

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