Have catch block return statement exit from parent function as well

1

Hello, I’m working with async, await and promises from within nodejs/express and I have this issue: In the catch block after my await, I want the return to also cause the parent function (in this case it’s a GET handler function) to return as well. However it doesn’t during a rejected promise from await, and code still executes as such:

app.route('/theRoute').get(function (req, res) {

  let result = await getPromiseThatRejects(data)
    .catch((error) => {
      res.send(error);
      return;
    })

 //additional code that I don't want executed if promise got error
  console.log("In the case of an error, this code still executes when I don't want it to")
})

This is the only solution I found:

app.route('/theRoute').get(function (req, res) {

  let success = true
  let result = await getPromiseThatRejects(data)
    .catch((error) => {
      success = false
      res.send(error); 
    })

  if (!success) return
  console.log("Code no longer executes if there was an error, which is my goal")
})

So basically I just have to check the boolean after the await is done and return if it’s false. But is there a more eloquent way to do this without having to use extra lines for boolean checks, or am I stuck with this technique for now.

2

This isn’t correct syntax for async await, the handler has to use the async keyword. And you’re mixing the two syntaxes, which isn’t an issue, it’s just a bit confusing

app.route('/theRoute').get(async (req, res) => {
  try {
    let result = await getPromiseThatRejects();
    res.doSomething(result) // ??
  } catch (error) {
    return next(error); // ?? Something like that, pass on to error handler
  }
});
3

Apologies, my app has async in the get function but I forgot to include it in this example. Yes, that way works. Heh, I was doing it my way because I didn’t like all the extra brackets from putting in the try block, but that works better. Thanks.