So, from code below im getting array of udefined but to be honest idk why and how to fix it...
const promises = sessionsIds.map((id) =>
stripe.checkout.sessions
.listLineItems(id, { limit: 100 })
.then((err, lineItems) => {
return lineItems;
})
);
const usernames = await Promise.all(promises);
I have also tried with async await but the result is still the same.
Assuming that stripe.checkout.sessions.listLineItems(id, { limit: 100 }) is returning a regular, standard promise, then the problem is that you're treating is like a regular asynchronous callback with (err, value) as the functions arguments. That's not how .then() works. The first argument to .then() is a function that will be called with one argument, the resolved value. In fact, you probably don't even need a .then() here.
You can change to this:
const promises = sessionsIds.map((id) =>
stripe.checkout.sessions.listLineItems(id, { limit: 100 })
);
const usernames = await Promise.all(promises);
Or, simplifying further:
const usernames = await Promise.all(sessionsIds.map((id) =>
stripe.checkout.sessions.listLineItems(id, { limit: 100 })
));
Note, the .then() handler here is not necessary (assuming stripe.checkout.sessions.listLineItems(id, { limit: 100 }) already returns a promise). If you want to use a .then() handler, it would go like this:
const promises = sessionsIds.map((id) =>
stripe.checkout.sessions.listLineItems(id, { limit: 100 })
.then((lineItems) => {
console.log(lineItems);
return lineItems;
});
);
const usernames = await Promise.all(promises);
Where you pass it a function as the first argument and it gets one argument (the resolved value).
Collected from the Internet
Please contact [email protected] to delete if infringement.
Comments