Question

谁能告诉我我将一个函数定义为异步函数，但仍然无法正常工作并引发错误。我该如何解决此问题？

var courses = [
    {ID: 1, Name: 'course 1', Author: 'Mariela Mckinney', Website: 'www.loarpharei.com'},
    {ID: 2, Name: 'course 2', Author: 'Brylee Becker', Website: 'www.phattiodye.com'},
    {ID: 3, Name: 'course 3', Author: 'Weston Edwards', Website: 'www.pretathlew.com'},
    {ID: 4, Name: 'course 4', Author: 'Nathanael Fox', Website: 'www.senfredral.com'},
    {ID: 5, Name: 'course 5', Author: 'Aarav Park', Website: 'www.ebreanerin.com'}
]

async function get_course(ID){
    return setTimeout(function(){
        console.log('I am fired');
        return get_index(ID);
    }, 2000);
}

let course_id = await get_course(2);
console.log(course_id);

Answer 1

代码中有很多错误。希望代码段对您有所帮助。

const courses = [
  {id: 1, name: 'course 1', author: 'Mariela Mckinney', website: 'www.loarpharei.com'},
  {id: 2, name: 'course 2', author: 'Brylee Becker', website: 'www.phattiodye.com'},
  {id: 3, name: 'course 3', author: 'Weston Edwards', website: 'www.pretathlew.com'},
  {id: 4, name: 'course 4', author: 'Nathanael Fox', website: 'www.senfredral.com'},
  {id: 5, name: 'course 5', author: 'Aarav Park', website: 'www.ebreanerin.com'},
]

// The "async" keyword is actually redundant if there is no "await" keyword in
// the function.
async function getCourse(id) {
  return new Promise((resolve) => {
    setTimeout(() => {
      console.log('I am fired');
      resolve(courses[id]);
    }, 2000);
  });
}

async function main() {
  // You can always "await" an Async function, or a function returning a Promise.
  const course = await getCourse(2);
  console.log(course);
}

main();

// Notes:
// 1. Always use camelCase for field names, function names and variable names.
// 2. Most JavaScript developers prefer 2-space indent, instead of 4.

Answer 2

您不能在await函数之外使用async。 API的设计方式。请参阅https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/await上的文档：

await运算符用于等待Promise。它只能在异步函数中使用。

如果要使用await，则需要包装

let course_id = await get_course(2);

在async函数内部。

而且，正如丹尼尔·A·怀特（Daniel A. White）所指出的那样，setTimeout不会返回Promise，而是一个计时器ID，您以后可以使用它取消超时。

Answer 3

您可以使用IIFE将等待调用包装在async函数中，但还需要其他函数的返回承诺。

var courses = [{ID: 1, Name: 'course 1', Author: 'Mariela Mckinney', Website: 'www.loarpharei.com'},{ID: 2, Name: 'course 2', Author: 'Brylee Becker', Website: 'www.phattiodye.com'},{ID: 3, Name: 'course 3', Author: 'Weston Edwards', Website: 'www.pretathlew.com'},{ID: 4, Name: 'course 4', Author: 'Nathanael Fox', Website: 'www.senfredral.com'},{ID: 5, Name: 'course 5', Author: 'Aarav Park', Website: 'www.ebreanerin.com'}]

function get_course(ID) {
  console.log('I am fired.')
  return new Promise(resolve => {
    setTimeout(() => {
      resolve(courses)
    }, 2000)
  })
}

(async() => {
  let course_id = await get_course(2);
  console.log(course_id);
})()

Answer 4

setTimeout不返回承诺。您必须将其包装在new Promise(...)调用中。 get_course(2)只是一个自由浮动的调用，它不在任何函数中，因此await不会在其中起作用。

Answer 5

您应该将setTimeout包装到Promise中。参见示例：

 (async() => {var courses = [
   {ID: 1, Name: 'course 1', Author: 'Mariela Mckinney', Website: 'www.loarpharei.com'},
   {ID: 2, Name: 'course 2', Author: 'Brylee Becker', Website: 'www.phattiodye.com'},
   {ID: 3, Name: 'course 3', Author: 'Weston Edwards', Website: 'www.pretathlew.com'},
   {ID: 4, Name: 'course 4', Author: 'Nathanael Fox', Website: 'www.senfredral.com'},
   {ID: 5, Name: 'course 5', Author: 'Aarav Park', Website: 'www.ebreanerin.com'}
 ]
 
 function get_index(id) {
 		return courses.findIndex(i => i.ID === id);
 }

 function get_course(ID){
   return new Promise(resolve => {
     setTimeout(function(){
       console.log('I am fired');
        resolve(get_index(ID));
   }, 2000);
   });
 }

 let course_id = await get_course(2);
 console.log(course_id);
})();

JS仅在异步功能中有效。怎么解决？

5 个答案: