correctly handle async/await on a mysql query

Question

I am trying to perform a mysql query asyncronously assigning a value to a const to be later used in my code.

What I have so far is:

const plan_free = await con1.promise().query("SELECT stripe_price1 FROM funzioni WHERE titolo='SmartBanca'")
.then( ([rows,fields]) => {
    return rows[0]['stripe_price1'];
})
.catch(console.log)
.then( () => con1.end());
console.log("Plan_free: "+ plan_free);

but plan_free returns [object Object]. I am sure I am close to the goal but I don't get what I am still doing wrong.

Edit: by adding the following lines:

plan_free1 = JSON.stringify(plan_free);
console.log("Plan_free: "+ plan_free1);

I see the following content for plan_free:

Plan_free: {"_events":{},"_eventsCount":0,"next":null}

that actually is not even close to what I want to achieve.

Isn't this the case, because the function is async and thus the line where you do the console.log is getting called before your async task might be finnished? — Olli, Jan 04 '21 at 15:25
Sure, your console log must be in the resolved then section : just the line following the line where plan_free is declared. — tsamaya, Jan 04 '21 at 15:26
I need to have the result after the call, not in the callback, in the callback it is correctly evaluated. So I need to fix it. See the edit. — Lelio Faieta, Jan 04 '21 at 15:47

score 0 · Accepted Answer · answered Jan 04 '21 at 16:23

Ok, I finally made it.

var plain_free='';
await con1.promise().query("SELECT stripe_price1 FROM funzioni WHERE titolo='SmartBanca'")
.then( ([rows,fields]) => {
    plan_free = rows[0]['stripe_price1'];
    return plan_free;
})
.catch(console.log)
.then( () => con1.end());
            
console.log("Plan_free: "+ plan_free);

This way I was able to fetch the value from the database and assign it to a variable in node. This will be later used in my code without the need to nest it in callback of the query.

correctly handle async/await on a mysql query

1 Answers1