如何解决我的异步功能？

Question

I want to get the number of users for each shop in my database. So I'm making two call in my Db, one for the shop and another for the users. And I have a problem I don't have anything in my "nbr_users" return value.. Could you help ?

function nbr_users(key) {
    return new Promise((resolve, reject) => {
        db.User.find({
            shop: key
        }, (err, users) => {
            if (err) reject(err)
            else {
                resolve(users.length) // Maybe my problem is here...
            }
        })
    })
}

module.exports = (req, res) => {
    db.Account.find({},
    (err, accounts) => {
        if (err) res.json(err)
        else {
            var accountMap = {}
            var i = 0;
            accounts.forEach(async function(account) {
                accountMap[i++] = {
                    users: await nbr_users(account.key) // the value I want
                }
              });
            res.json({user_list: accountMap})
        }
    })
}

Answer 1

You have something returning from your nbr_users(key) . You have a promise, then you may use .then .

Because you are in a loop with async. functions and you don't want to lose track of i I'd recomend you to use anonymous Immediately-invoked function expression .

function(i){
    nbr_users(account.key).then(function(length){
      accountMap[i] = {
        users: length
    })
}(i++);

Answer 2

accounts.forEach(async function(account) {
            accountMap[i++] = {
                users: await nbr_users(account.key) // the value I want
            }
          });

Should be more like

var accountMap = {}
var i = 0;
for (let account of accounts) {
   let users = await nbr_users(account.key);
   accountMap[i++] = {
                users: users
            }
}

res.json({user_list: accountMap});

Make sure you're running that in an async function.

Downside of this way is that each subsequent nbr_users(account.key) request must wait for the previous one. There is a way to run them all without each one waiting for the previous as well, let me know if you want me to show that way -- it's slightly more complicated