javascript-如何实现可取消的,有序的诺言？

我整理了一个例子来说明我要得到的是：

function onInput(ev) {
  let term = ev.target.value;
  console.log(`searching for "${term}"`);
  getSearchResults(term).then(results => {
    console.log(`results for "${term}"`,results);
  });
}

function getSearchResults(term) {
  return new Promise((resolve,reject) => {
    let timeout = getRandomIntInclusive(100,2000);
    setTimeout(() => {
       resolve([term.toLowerCase(), term.toupperCase()]);  
    }, timeout);
    
  });
}

function getRandomIntInclusive(min, max) {
  min = Math.ceil(min);
  max = Math.floor(max);
  return Math.floor(Math.random() * (max - min + 1)) + min;
}

<input onInput="onInput(event)">

在“搜索”框中键入并观察控制台.搜索结果重新排序！

当有新的输入时,我们如何取消任何未完成的承诺并保证结果按顺序返回？

解决方法:

我没有使用去抖动或超时,而是在使用引用函数的该函数的内部(Jaromanda X建议)设置了少量状态.这样,您可以将函数引用更改为类似noop的内容.承诺仍然可以解决,但不会采取任何行动.但是,最后一个不会更改其功能参考：

var onInput = function() {
  let logger = function(term, results) {
    console.log(`results for "${term}"`, results);
  };
  let noop = Function.prototype;
  let lastInstance = null;

  function ActionManager(action) {
    this.action = action;
  }

  return function onInput(ev) {
    let term = ev.target.value;
    console.log(`searching for "${term}"`);

    if (lastInstance) {
      lastInstance.action = noop;
    }

    let inst = new ActionManager(logger.bind(null, term));
    lastInstance = inst;

    getSearchResults(term).then(response => inst.action(response));
  }
}();



/****************************************
 * The rest of the JavaScript is included only for simulation purposes
 ****************************************/

function getSearchResults(term) {
  return new Promise((resolve, reject) => {
    let timeout = getRandomIntInclusive(100, 2000);
    setTimeout(() => {
      resolve([term.toLowerCase(), term.toupperCase()]);
    }, timeout);

  });
}

function getRandomIntInclusive(min, max) {
  min = Math.ceil(min);
  max = Math.floor(max);
  return Math.floor(Math.random() * (max - min + 1)) + min;
}

<input onInput="onInput(event)">