Database Querying function not running asynchronously

Question

So I created a function in my node server which takes in a query string, runs it on my db, and returns the results. I then wanted to use my function asynchronously using async await throughout my routes instead of having nested query within, nested query, within nested query etc.

So here is the code:

 const runQuery = queryString => {
    console.log("...runQuery")
    db.query(queryString, (error, results, fields) => {
        if (error) {
            console.log("runQuery: FAILURE");
            return error;
        }
        else {
            console.log("runQuery: SUCCESS");
            return(results);
        }
    })
}

register.post("/", async (req, res) => {
    console.log(req.body);
    const results = await runQuery("select * from test1");
    res.send(results);
})

The database should have 3 entries, but unfortunately, it returns nothing. Meaning results is an empty variable when it is sent, meaning JS never properly waits for it to capture the db results. How can I use my function asynchronously, and how is this even feasible?

its a basic mysql connection object i imported from another file. — Mahdiur Rahman
– Mahdiur Rahman, Commented Feb 23, 2020 at 7:01
I know it's a connection object. When people ask, we usually mean put the code as well or describe it like for example mysql from package some-pacakge. You're trying to use Promise but not all package provides promise support — 1565986223
– 1565986223, Commented Feb 23, 2020 at 7:23
Ahhh I see. Apologies about that. But yes I checked the problem again today and see that the mysql package doesn't have promise support. my db connection is from the regular mysql package. — Mahdiur Rahman
– Mahdiur Rahman, Commented Feb 23, 2020 at 23:20

marc_s · Accepted Answer · 2021-07-16 09:29:09Z

1

It seems your function "runQuery" does not return a promise, in fact, it's not returning anything. You are using "return" in the callback of the db.query function, not the function "runQuery" itself.

Since runQuery is performing an asynchronous operation, the result ought to be resolved via a promise (which is what the "await" in your request handler is looking for).

I'm not exactly sure but it seems you are using MySql, and I could not find anything on the npm page of the mysql package regarding the query being promisified, so we'll promisify it ourselves:

const runQuery = (queryString) => new Promise((resolve,reject) => {
    console.log("...runQuery")
    db.query(queryString, (error, results, fields) => {
        if (error) {
            console.error("runQuery: FAILURE");
            reject(error);
        } else {
            console.log("runQuery: SUCCESS");
            resolve(results);
        }
    })

})

register.post("/", async (req, res) => {
    console.log(req.body);
    try{
        const results = await runQuery("select * from test1");
        res.send(results);
    }catch(e){
        console.error(`ERROR THROWN BY runQuery:`,e);
        res.status(500).send({
            message: e.message || "INTERNAL SERVER ERROR"
        })
    }
})

Note that if an error occurs, our promisified function will reject the error and it will NOT be stored in the "results" variable in our request handler. Instead, an error will be thrown which needs to be handled. Therefore, it is always a good practice to put any async/await calls inside a try/catch.

edited Jul 16, 2021 at 9:29

marc_s

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answered Feb 23, 2020 at 7:08

Anuj Pancholi

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2 Comments

Mahdiur Rahman Over a year ago

Thanks so much Anuj, your answer worked and taught me something about promises! My apologies for not mentioning that I was using the basic mysql package. It turns out, as another user mentioned, that not all packages support promises. I assumed all db-management packages had promise wrappers automatically applied. I am an intern so this is all sort of new to me but your answer has helped show me a new dynamic way to use the Promise JS class. Thank so much. Unfortunately, my rep is too low to upvote, otherwise you woulda gotten one. Sorry :(

Anuj Pancholi Over a year ago

You're welcome! It's ok, we're all here to learn. Keep coding!

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