Question

这个承诺的位置我做错了什么？

我需要先完成第一个功能的所有元素，然后才能触发第二个功能。

谢谢。

功能1：

var promise1 = $("#button-1").on('click', async new Promise(function(resolve, reject) {

  csInterface.evalScript("Duplicate()");
  await SafeMode();
  csInterface.evalScript("FlattenImage()");
  await SafeMode();
  csInterface.evalScript("ConverttosRGB()");
  await SafeMode();
  csInterface.evalScript("Resolution300PPI()");
  await SafeMode();
  csInterface.evalScript("SetBackgroundtoLayer()");
  await SafeMode();
  csInterface.evalScript("Postcard1()");
  resolve();
}));

功能2：

promise1.then(function() {

// next function here

});

Answer 1

promise1不是一个承诺，实际上是一个jQuery对象。制作一个可以生成您所需的诺言的函数（见下文）。

const promise1 =() => new Promise(async function(resolve, reject) {

  /** Commented so you can see that it works
  csInterface.evalScript("Duplicate()");
  await SafeMode();
  csInterface.evalScript("FlattenImage()");
  await SafeMode();
  csInterface.evalScript("ConverttosRGB()");
  await SafeMode();
  csInterface.evalScript("Resolution300PPI()");
  await SafeMode();
  csInterface.evalScript("SetBackgroundtoLayer()");
  await SafeMode();
  csInterface.evalScript("Postcard1()");*/
  
  //Adding time
  setTimeout(()=>{
    console.log("promise1");
    resolve();
  }, 500);
  
})

$("#button-1").on('click', () => {
  promise1().then(() => {
    //next bit of your code here
    console.log("next");
  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<button id="button-1">Click me</button>

Answer 2

on()函数不会返回promise，您需要先声明异步函数，然后将其作为回调回调并与THEN链接起来。

此外，异步功能将自动返回Promise，而无需您自己构建。

const resolveAfter1Seconds = function(id) {
  return new Promise(resolve => {
    setTimeout(function() {
      resolve("resolution " + id);
    }, 1000);
  });
};

const callback = async function() {

  console.log("in callback", "start");
  
  let res = await resolveAfter1Seconds(1);
  console.log("in callback 1", res);
  res = await resolveAfter1Seconds(2);
  console.log("in callback 2", res);
  res = await resolveAfter1Seconds(3);
  console.log("in callback 3", res);
  
  return "what you want your promise to finally return";
};


$("#button-1").on('click', () => {
  callback().then(x => console.log("THEN", x))
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<button id="button-1">Click me!</button>

Javascript Promise的放置问题

2 个答案: