Question

我有一个async函数，需要返回一个user_ids数组。我的代码就像：

async function getUserIds(option) {
  let user_ids = [];

  if (option === 'test') {
    const departments = await Departments.find({}).exec()
      .catch(() => console.log('ERROR');

    console.log('BEFORE LOOP');

    await departments.forEach(async(dpt) => {
      const pop = await Populations.findOne({_id: dpt.id}) 
        .catch(() => console.log('ERROR');

      console.log('work...');

      if (pop && (pop.max > pop.min)) {    
        const _users = await User.find({_id: {$in: pop.ids}}).exec()
          .catch(() => console.log("ERROR");

        user_ids = user_ids.concat(_users.map((u) => u._id));
        console.log('finished work...');
      }
    });

    return user_ids;
}

async function main() {
    let user_ids = await getUserIds('test');
    console.log(user_ids);
}

现在，这总是返回一个空数组[]，我可以以异步方式看到控制台日志：

BEFORE LOOP
[]   --> the return 
work...
work...
work...
work...
work...
work...
finished work...
finished work...
work...
finished work...
work...
finished work...
finished work...

我猜这行await departments.forEach(async(dpt) => {并不是真的"awaiting"，所以我该怎么办，我做错了什么？

Answer 1

async / await是使用Promise的一种糖方式。

async就像new Promise()，而await就像.then()。

使用Promise.all()等到所有Promise的决心都解决后再继续。

在这种情况下，您应该使用Array.prototype.map()而不是Array.prototype.forEach()。 forEach()返回undefined，您正在使用await。 .map()将返回一个Promise数组：

// Using .map()
departments = [ Promise, Promise, Promise ] // await till all Promise resolves to go foward.

// Using .forEach()
departments = 'undefined' // nothing to await for.

await Promise.all(
        departments.map(async(dpt) => {
            const pop = await Populations.findOne({_id: dpt.id})
                .catch(() => console.log('ERROR');

            console.log('work...');

            if (pop && (pop.max > pop.min)) {    
                const _users = await User.find({_id: {$in: pop.ids}}).exec()
                    .catch(() => console.log("ERROR");

                user_ids = user_ids.concat(_users.map((u) => u._id));
                
                console.log('finished work...');
            }
       })
);

Answer 2

在从函数返回结果之前，您应该等待部门的承诺。第forEach行在您的情况下不起作用，请用map或for of

代替

以map为例

...
const promises = departments.map(async(dpt) => {
  const pop = await Populations.findOne({_id: dpt.id}) 
    .catch(() => console.log('ERROR');

  console.log('work...');

  if (pop && (pop.max > pop.min)) {    
    const _users = await User.find({_id: {$in: pop.ids}}).exec()
      .catch(() => console.log("ERROR");

    user_ids = user_ids.concat(_users.map((u) => u._id));
    console.log('finished work...');
  }
});

await Promise.all(promises)

return user_ids

嵌套异步/等待猫鼬查询

2 个答案: