CODEWITHSUNDEEP
javascriptasync-await
Alexander Craggs
Alexander Craggs

Reputation: 8759

When using async/await, how do you stop execution of a function when one call errors out?

Lets imagine we have a function as so:

Router.get('/', async function (req, res, next) {
  let result = await someFunction().catch(next)
  someOtherCall()
})

If this errors out, it continues on to the global error handler by calling next(error_reason). However, if the someFunction() fails, we don't want someOtherCall() to run at all. At the moment, I can see two ways of fixing this:

// Suggestion from https://stackoverflow.com/q/28835780/3832377
Router.get('/', async function (req, res, next) {
  let result = await someFunction().catch(next)
  if (!result) return // Ugly, especially if we have to run it after every call.
  someOtherCall()
})

Router.get('/', async function (req, res, next) {
  let result = someFunction().then(() => {
    // Much less work, but gets us back to using callbacks, which async/await were
    // meant to help fix for us.
    someOtherCall()
  }).catch(next)
})

Is there a simpler way to stop a function from executing if any of the functions call that doesn't mean adding another statement after every function call or using callbacks?

Upvotes: 2

Views: 1416

Answers (1)

Alberto Trindade Tavares
Alberto Trindade Tavares

Reputation: 10356

You can simply use try-catch:

Router.get('/', async function (req, res, next) {
  try { 
    let result = await someFunction()
    someOtherCall()
  }
  catch(exception) {
    next(exception)
  }
})

Upvotes: 3

Related Questions