Ignoring exception from a finished function on Node JS

Question

I am trying to delete votes from my firebase db table and from my understanding it is an async request that takes time, though my function ends way before than so I get the following error -

Ignoring exception from a finished function

here is my function code -

function continueDeleteFromFirebaseVotesDB(videoId, contestId) {
    console.log("before query declaration");
    var query = database.ref("votes/" + contestId).orderByKey();
    console.log("before foreach loop")
    query.once(value)
        .then(function (snapshot) {
            console.log("inside foreach loop")
            if (!snapshot.exists) {
                console.log("votes db snapshot does not exist");
                return;
            }
            snapshot.forEach(function (childSnapshot) {
                //var key = childSnapshot.key;
                // childData will be the actual contents of the child
                var childData = childSnapshot.val();
                if (childData.video_id === contestId) {
                    childSnapshot.ref.remove();
                    console.log("removed vote - " + JSON.stringify(childSnapshot))
                    continueDeleteFromFirebaseVideosDB(videoId)
                }
            });
            return query;
        })
        .then(function (data) {
            console.log(JSON.stringify(data));
        })
}

What am I missing in order to make my function async and wait for the result?

array.forEach() doesn't handle async functions. Try taking a look at this question https://stackoverflow.com/questions/37576685/using-async-await-with-a-foreach-loop — VladNeacsu, Sep 12 '19 at 12:23
Why are you using `continueDeleteFromFirebaseVideosDB(videoId)` after the removing operation ? — Emeric, Sep 12 '19 at 14:10

score 1 · Answer 1 · answered Sep 12 '19 at 16:12

query.once(value) returns immediately with a promise, even though you call then on that promise. That means your function returns immediately, and the query finishes some time later. If your function performs async work, it should return a promise, and the caller should use that promise to collect any results that it contains.

You can't simply make async work asynchronous, if you were thinking about trying to make your function block until the query is complete. All promises must be handled correctly.

score 0 · Accepted Answer · answered Sep 15 '19 at 07:22

0

Found the answer. I needed to write "value" and not value.

answered Sep 15 '19 at 07:22

Alon Shlider

1,187
1
16
46

Ignoring exception from a finished function on Node JS

2 Answers2