CODEWITHSUNDEEP
typescriptasync-await
tmp dev
tmp dev

Reputation: 9193

async await, continue without the catch

I have the following code

  try {
    const result = await doSomething();
  }
  catch {
    console.log('Error loaidng rejected data');
  }

  // rest of code

My problem is that I'm not doing anything in the catch segment. I'm only including it so that the code continues in case there is an error. If I remove the catch then the whole thing blows up. Is there anyway I can achieve the same without a catch? and without using promises?

Upvotes: 3

Views: 3159

Answers (3)

Raja Jaganathan
Raja Jaganathan

Reputation: 36117

Yes, most of the time try/catch block is huge, to make it concise you can leverage await-to-js libs. So we can use like below.

const [err, result] = await to(doSomething());

Note that it never blows up because we attached catch handler for this promise.

I just copy the source from https://github.com/scopsy/await-to-js.

export function to<T, U = Error> (
  promise: Promise<T>,
  errorExt?: object
): Promise<[U | null, T | undefined]> {
  return promise
    .then<[null, T]>((data: T) => [null, data])
    .catch<[U, undefined]>((err: U) => {
      if (errorExt) {
        Object.assign(err, errorExt);
      }

      return [err, undefined];
    });
}

export default to;

Upvotes: 0

hoangdv
hoangdv

Reputation: 16127

It is a Promise, then you can ignore the catch function.

const result = await doSomething().catch(() => null);

Do nothing in error case, your code will continue execute next line, but result is undefined in this case (error case).

Upvotes: 9

Adrian Brand
Adrian Brand

Reputation: 21628

Can I do it without a promise?

No, await is just syntactic sugar to wrap a promise. If your async function does not return a promise then it doesn't make sense to be using await.

You could catch the error inside doSomething and resolve the promise with null or undefined to keep things going.

Upvotes: 2

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