Q承诺的串行执行

Question

我想我误解了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:

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

我究竟做错了什么？

Answer 1

这有效:

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

这是有道理的 - 只是doWork直接调用then()将立即触发超时,而不是让Q有机会等到task one完成.