How to use result of called 'then' for next 'then'?

Question

I want to return what I get after an async call.

So

app.get("/books", async (req, res) => {
    let books = await getBooks()
        .then(json => {
            res.status(200).send({"books": json});
        }); 
});

Should wait on rendering the result until the called getBooks is done.

export async function getBooks() {
    console.log("Getting books from cloud");
    Book.findAll({
        // ...

    }).then(books => {
        console.log("Got books");
        return JSON.stringify(books, null, 4);
    });
}

But right now the response gets rendered without actually waiting for the result.

Check this: https://javascript.info/promise-chaining or this https://stackoverflow.com/questions/14220321/how-do-i-return-the-response-from-an-asynchronous-call It's probably gonna help you. — Naor Levi, Jan 23 '20 at 07:29

score 2 · Answer 1 · answered Jan 22 '20 at 19:11

You don't need to use promises. You can just use await and then use the result of that.

 app.get("/books", async (req, res) => {
   const books = await getBooks();
   res.status(200).send({ books });
 });

I'd highly suggest taking it a step further and using try/catch to handle failure cases

 app.get("/books", async (req, res) => {
   try {
     const books = await getBooks();
     res.status(200).send({ books });
   } catch (error) {
     // massage this to send the correct status code and body
     res.status(400).send( { error });
   }
 });

Correct and my answer was somewhat misleading. I know async/await uses promises under the hood but I didn't want to get into that deep of a conversation for this answer :D — VtoCorleone, Jan 24 '20 at 04:19

score 0 · Answer 2 · answered Jan 22 '20 at 19:08

0

You can use await in second method too:

export async function getBooks() {
console.log("Getting books from cloud");
var books = await Book.findAll({
    // ...

})
 if(books){
     console.log("Got books");
    return JSON.stringify(books, null, 4);
  }

answered Jan 22 '20 at 19:08

Juan Andres solanas

185
7

score 0 · Answer 3 · answered Jan 22 '20 at 19:08

You just need to return your promise, and since you're just returning a promise you don't need async;.

app.get("/books", async (req, res) => {
    let json = await getBooks()
    res.status(200).send({"books": json});
});

export function getBooks() {
    console.log("Getting books from cloud");
    return Book.findAll({
        // ...

    }).then(books => {
        console.log("Got books");
        return JSON.stringify(books, null, 4);
    });
}

score 0 · Answer 4 · edited Jan 23 '20 at 06:45

0

app.get("/books", async (req, res) => {
    let books = await getBooks();
    res.status(200).send({"books": books});
})

edited Jan 23 '20 at 06:45

Nandita Sharma

13,287
2
22
35

answered Jan 22 '20 at 19:10

Dilip Kale

71
7

How to use result of called 'then' for next 'then'?

4 Answers4