javascript - How to get returned value from async generator when using for await - Stack Overflow

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I have following code:

for await (const part of resolveData(data)) {
    console.log({part});
}

It iterates over an async generator that looks like this:

const resolveData = async function* <T> (data: SomeValue<T>) {
    if (<expression1>) {
        if (<expression2>) {
            console.log("valid");
            yield 1;
        }
    }
    return 2;
}

The output of the for await loop looks like this:

{ part: 1 }

I want to get the returned value (number 2 in this case) as well. How to get the value inside or outside of the loop?

Tried looking online but found nothing.

I have following code:

for await (const part of resolveData(data)) {
    console.log({part});
}

It iterates over an async generator that looks like this:

const resolveData = async function* <T> (data: SomeValue<T>) {
    if (<expression1>) {
        if (<expression2>) {
            console.log("valid");
            yield 1;
        }
    }
    return 2;
}

The output of the for await loop looks like this:

{ part: 1 }

I want to get the returned value (number 2 in this case) as well. How to get the value inside or outside of the loop?

Tried looking online but found nothing.

• javascript
• typescript
• ecmascript-6

Share Improve this question Follow edited Dec 28, 2023 at 15:13 sss3243 asked Dec 28, 2023 at 15:09 sss3243sss3243 9555 bronze badges 1

• 1 This is a typo. I meant to create only one generator – sss3243 Commented Dec 28, 2023 at 15:12

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 8

When you use return in a generator or async generator, that sets the value of value on the result object for when done is true. But for-of and for-await-of don't run the body of the loop for the result object where done is true. Here's a simpler example:

<script type="module">
// Note: Using an async iterator even though we never await
// because the OP's code is async
async function* example() {
    yield 1;
    return 2;
}

for await (const x of example()) {
    console.log(x);
}
</script>

(Sadly, the "JavaScript" panel in Stack Snippets can't be run as a module, and I wanted to use top-level await, so I had to put the JavaScript in the "HTML" panel instead.)

Notice how only 1 is shown. (The same is true for for-await-of.)

You have at least two options:

1. Change your return 2 to yield 2, so that the generator isn't "done" yet as of when it yields 2, like this:

<script type="module">
// Note: Using an async iterator even though we never await
// because the OP's code is async
async function* example() {
    yield 1;
    yield 2; // `yield` instead of `return`
}

for await (const x of example()) {
    console.log(x);
}
</script>

2. Change your code to use the iterator directly rather than using for-await-of.

<script type="module">
   // Note: Using an async iterator even though we never await
   // because the OP's code is async
   async function* example() {
       yield 1;
       return 2;
   }
   
   const it = example();
   let r;
   while (!(r = await it.next()).done) {
       console.log(`Value = ${r.value}, not done yet`);
   }
   console.log(`Value = ${r.value}, done`);
   </script>

In both cases, you now see 1 and 2.

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