稍后解决承诺

ale*_*kop 5 es6-promise

我想构建一个 Promise,但将解决方案推迟到以后。下面的代码创建了一个承诺,但它立即得到解决。我如何控制承诺何时被评估?

var p = new Promise((resolve, reject) => {
        resolve(1);
    })
    .then((p1) => {
        console.log(p1 + 1);
    });
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更新:为了澄清,想要将承诺的声明与其执行分开的原因是then基于一些参数动态添加回调。

Lou*_*uis 5

您可以将resolve和传递reject给您想要使用的任何异步函数。并且这样的函数可以在完成其工作时调用它。这是一个在 Node 中运行的示例。如果您运行它,它将ls -l在您的当前目录中执行。该execSomething函数仅接受回调,并且该promiseToExec函数将resolve, reject回调传递给execSomething而不是立即调用它们中的任何一个。

const childProcess = require("child_process");

function execSomething(command, options, onSuccess, onError) {
  childProcess.exec(command, options, (err, stdout, stderr) => {
    if (err) {
      onError(err);
    }
    onSuccess(stdout, stderr);
  });
}

function promiseToExec(command, options) {
  return new Promise((resolve, reject) => {
      execSomething(command, options, resolve, reject);
  });
}

promiseToExec("ls -l").then(console.log.bind(console));
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卡兹劳斯基斯建议这样做:

var resolve;
var promise = new Promise(function(fulfill) {
  resolve = fulfill;
});
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不要这样做!

当您传递给的回调中发生异常时new Promise,promise 的规范是异常将自动转换为 Promise 拒绝。因此,如果throw Error...回调内发生任何事情,您将获得自动转换。

如果您保存resolve回调并将逻辑移至传递给的回调之外new Promise,则您不会获得此自动转换。回调外部抛出的异常只会在堆栈中向上传递,而不会转换为承诺拒绝。这很糟糕,因为它要求函数的用户用来.catch捕获被拒绝的承诺 try...catch抛出的异常。这是一种糟糕的设计实践。

这是说明问题的代码:

// This is how things should be done.
function makeGoodPromise(num) {
  return new Promise((resolve) => {
    if (num < 0) {
      throw new Error("negative num");
    }
    resolve(num);
  });
}

// This is a bad approach because it will sometimes result in synchronous
// exceptions.
function makeBadPromise(num) {
  let resolve;
  const p = new Promise((fullfil) => {
    resolve = fullfil;
  });

  if (num < 0) {
    throw new Error("negative num");
  }
  resolve(num);

  return p;
}

// Shoring up the bad approach with a try... catch clause. This illustrates what
// you need to do convert the exception into a rejection. However, why deal with the
// additional scaffolding when you can just take the simpler approach of not
// leaking the callbacks??
function makeBadPromise2(num) {
  let resolve, reject;
  const p = new Promise((fullfil, deny) => {
    resolve = fullfil;
    reject = deny;
  });

  try {
    if (num < 0) {
      throw new Error("negative num");
    }
    resolve(num);
  }
  catch (e) {
    reject(e);
  }

  return p;
}


makeGoodPromise(-1).catch(() => console.log("caught the good one!"));

try  {
  makeBadPromise(-1).catch(() => console.log("caught the bad one!"));
}
catch(e) {
  console.log("Oops! Synchronous exception: ", e);
}

makeBadPromise2(-1).catch(() => console.log("caught the bad2 one!"));
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当我在 Node 中执行它时,这是输出:

Oops! Synchronous exception:  Error: negative num
    at makeBadPromise (/tmp/t12/test2.js:17:11)
    at Object.<anonymous> (/tmp/t12/test2.js:48:3)
    at Module._compile (module.js:570:32)
    at Object.Module._extensions..js (module.js:579:10)
    at Module.load (module.js:487:32)
    at tryModuleLoad (module.js:446:12)
    at Function.Module._load (module.js:438:3)
    at Module.runMain (module.js:604:10)
    at run (bootstrap_node.js:394:7)
    at startup (bootstrap_node.js:149:9)
caught the good one!
caught the bad2 one!
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Ste*_* H. 1

不完全确定您在问什么 - 下面的代码演示了承诺的构造和调用then以相同的顺序发生,但可能在不同的时间执行。wait1更改和的值wait2并查看输出有何不同,但无论时间如何,代码都可以正常工作。

我认为由您来实现承诺代码,以便它等待您想要等待的任何条件。在示例中,它是一个简单的setTimeout,但可以想象您可以采取任何措施来推迟执行。

您可能需要使用 Chrome 才能在浏览器中查看这些结果:

var wait1 = 2000;
var wait2 = 1000;

function log(...str) {
  var li = document.createElement('li');
  str.forEach((s) => {
    li.appendChild(document.createTextNode(s));
  });
  document.getElementById("log").appendChild(li);
}

var p = new Promise((resolve, reject) => {
  setTimeout(() => {
    log("Resolving promise!");
    resolve(1);
  }, wait1);
});

log("Promise created!");

setTimeout(() => {
  log("Calling 'then'");
  p.then((p1) => {
    log("Value:", p1 + 1);
  });
}, wait2);
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<ol id="log" />
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